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The solution of homework question is Let Z=a + ib So Z conjugate = a-bi Now Z²=Z conjugate a² + b²i² + 2aib = a-bi a² - b² +2aib = a-bi Now by comparing the real part by real and imaginary part by imaginary a = a² - b² b = -2aib a = -1/2 Putting this in a = a² - b² we get b = +(√3/2), (-√3/2) So complex number Z=(-1/2) +(√3/2) i and Z = (-1/2) -(√3/2) i
Salute sir aapne yah lec 3 baar record kiya sirf hamare liye sir aap logo ki wajah se itna achcha quality education free milta hai aur sir yah aap logo ki mehnat hamare result mein chamkegi
36:49 Answer Omega, 0 and 1 Why because if we consider the Z as a+ib, then we have got a imaginary value i.e (-1+i√3) devided by 2 = omega And the other values should 0 and 1 By putting we got
Given That:- z^2 =zconjugate To Find- Possible Solution Of Z Soln - (Step1):- write (Z^2 = Z conjugate) in modulus. :- |Z|^2 = |Z conjugate| ..by Modulus Property ( |Z Conjugate| =|Z|) therefore, :- |Z|^2=|Z| ..(equation) (Step2)by Observing the above equation | Z | = 0&1 (Case1)-Put |Z|=0 then Z=0 ..Ans.1 (Case2)- Put |Z| =1 then (Z^2)=(Z conjugate) = (Z^-1)..ans.2 Z^3 =1 so,Z =1...ans.3 Therefore, this gives remaining 3 Solution .Hence |Z|^2 =|Z Conjugate| have over all 4 Solutions of 'Z' ❤️PHYSICSWALLAH ❤️PACE ❤️AMANSIR
All possible solution of z square =conjugate of z is . 1; if y=0 then x=0 2;if y=0 then x=1 3;if x=-1/2 then y=+√3/2 4;if x=-1/2 then y=-√3/2. Sir please check my answer. It is right or wrong.
Sir i really respect the hard work and determination you put in your lectures. I am highly thankful to you sir. Thank you so much Alakh sir for finding such gems.
Sir you cannot imagine what is the value of your teaching in our life. You teach mathematics like never before. We all respect you and your hardworking towards us .
Sir I had written solution of questions (Z^2 = Z conjugate) ..... possible solu But I will again write the solution 0,1, -1/2 + root 3/2i , -1/2 - root 3/ 2i. Solve these equation ; ; ; ; ; ; ; x^2 - y^2 = x............................... x^2 + y^2 = x............................... And this property /Z/^2 = Z. Z conjugate. So
*sir* *homework* *question* *answer* *is* Z=1 Solution Given Z^2=Z' Multiplying both sides by Z (Z)(Z^2)=(Z)(Z') Now R.H.S will equal to |Z|^2 So (Z)(Z^2)=|Z|^2 Z= 1
Homework: 4solutions (0+i0), (1+i0), (-1/2+i√3/2),(-1/2-i√3/2) Putting z=x+iy in the equation and then equating Re(z) and Im (z) we get 2 values of x=-1/2 and 0 and 3 values of y=±√3/2,0
bro bilkul shi hemera bhi yehi aya . pata ni sablo properties learning ke bare me comment kar rahe he parkoi ans ni comment karta. bas match karana tha . thanks
@@yearsago-wz3jo assume z equals to a+bi,and then conjugate is a-bi z ,squaring the value of z and put it equal to value of conjugate then compare them u will find a and b.
Sir, Last me question ka answer 4 solution hoga. Z² =z bar {|z bar|=|z|} Z²=|z| |Z|=0,1 Case1: |z|=0,z=0 Case2: |z|=1 Z²=zbar =z-¹ Z³= 1 So, total possible solutions = 4
I was watching this video on PW App. But after listening, the story of recording lectures 3 times, I come here for give this video a huge like... And a lot of respect to AMAN SIR. Bhannat Maths Teacher.
Answer to the question u gave in H.W Q. z^2=conjugate of z Let z=a+bi Then. z^2=a^2+2abi-b^2 Conjugate of z=a-bi A.T.Q a^2+2abi-b^2 =a - bi (a^2-b^2)+(2ab+b)I=a+0i Equating real and imaginary part, a^2-b^2=a ..1.. 2ab+b=0 a=-1/2 Putting value of a in ..1.. ,we get b^2=3/4 b=+√3/2 or -√3/2; z=-1/2+√3/2i z=-1/2-√3/2i Values of z : z= -1/2 +√3/2i and -1/2-√3/2i
let z = x + iy then z(conjugate) = x - iy and z^2 = x^2 - y ^2 +2ixy comparing we get, x = x^2 - y^2 and y = -2xy if y = 0 , then x = 0,1 if y not equal to zero then x = -1/2 solving for y we get y = [+ or - root(3) / 2 ]
H.W. Solution-z=0,1,-1/2+√ 3/2 i,-1/2-√ 3/2 i Just take modulus on both sides to get |z|=0,1 then multiply both sides by z to get z³=1 hence other solution are root of unity (found 1 and solutions of x²+x+1=0 by quadratic formula)
Really sir, you are very great. Hamaare liye ap kitna hard work karte ho. We promise, hum apka mehnat ko bekar hone nenhi denge. We will also do hard work with you to be successful. Thank you❤ sir🥰 ❤❤❤❤❤❤❤❤❤
Solution of last question is- z^2=zbar (a+ib)^2=a+in a^2-b^2+2abi=a-ib Case-1 ,2abi=-bi a=-1/2 Case-2. ,a^2-b^2=a ......1. now put the value of a in equation-1 Then we get, (1/2)^2+1/2=b^2 1/4+1/2=b^2 b=+(√3/2)and-(√3/2) Then Z= (-1/2+√3/2i )and( -1/2-√3/2) Now the number of solutions is 2
Homework question: Z^2 = Z (bar) |Z^2|=|Z| {|Z(bar)| = |Z|} On solving we'll get two cases. Case 1:- |Z|=0 then Z=0 Case 2:- |Z|=1 then Z^2 = (|Z|^2)/Z {Z(bar) = (|Z|^2/Z)} So, Z^3 = 1 And cuberoot of Z will give cube roots of unity that are 1, omega, omega^2. So there can be 4 values of Z in equation Z^2 = Z(bar) . They are 0, 1, omega and omega^2. Thank you!!😊
36:17 Sir we can assume z=a+bi So we will get a²-b²+2abi=a-bi So 2ab=-b.......equation 1 a=-1/2(for b≠0) Now a²-b²=a....... equation 2 Put a=-1/2 We will get b= ±√3/4 So z=-1/2+√3/2 z=-1/2-√3/2 Sir these both complex numbers are also cube root of unity represented as w and w² respectively where w²=w' Getting back to the question😜 Sir two more answers are possible..... From equation 1 For b=0 Equation 2 a²-0²=a a=0,+1 z=0+0i z=1+0i So finally there are 4 possible values of z Sir my name is Shashank
HW question answer 4 possible value 1 , 0, -1/2+rootunder3/4i , -1/2-rootunder3/4i explanation:- (a+bi)^2 = a- bi a^2 + b^2 +2abi = a - bi ...........(1) compare imaginary part 2ab = -b b+ 2ab=0 b(2a+1)=0 so b=0 or a=-1/2 now put the above value in real part of (1) you will get two value corresponding above a and b each so finally you will get 4 possible value of z
Given:- z2=z to find the no. of solution of z2=z Solution:- Given equation, z2=z Taking modulus both sides we get ∣z∣2=∣z∣{∣z∣=∣z∣} ∣z∣2=∣z∣ So,∣z∣=0,1 case(1),∣z∣=0,z=0 case(2),∣z∣=1, Then the equation is z2=z=z−1 z3=1
Solution of last question is- z^2=zbar (a+ib)^2=a+in a^2-b^2+2abi=a-ib Case-1 ,2abi=-bi a=-1/2 Case-2. ,a^2-b^2=a ......1. now put the value of a in equation-1 Then we get, (1/2)^2+1/2=b^2 1/4+1/2=b^2 b=+(√3/2)and-(√3/2) Then Z= (-1/2+√3/2i )and( -1/2-√3/2) Now the number of solutions is 2 ✌✌✌✌✌✌✌✌✌✌✌✌
Last question answer: z²=z bar taking modulus on both sides, |z²|=|z bar| Now, |z|=|z bar| =>|z²|=|z| =>|z|=0,1 |z|=0 is possible iff z=0 |z|=1 Taking square on both sides, z×z bar=1 =>z bar=1/z Also, z²=z bar =>z²=1/z =>z³=1 z³-1=0 We can see one solution is z=1 =>(z-1)(z²+z+1)=0 Solving, we get z=1,(-1/2)+-(√3 iota/2) =>there are 4 solutions: z=0, 1, (-1/2)+(√3 iota/2), (-1/2)-(√3 iota/2)
For bieng max value z1 should be equal to 1 or otherwise it will have smaller one then:- Z1-1=1 Z1=2 Same with z2 and z3 so:- Z2=4 Z3=6 And adding all will sum up 12 bhannat😀
36:40 my answer is coming 1 But I think that all the real number should satisfy the equation And also the equation given in question will be purely real
For that last question .... So for greatest value |z1+z2+z3| must be =|z1|+|z2|+|z3| Now a/c to ques. |z1-1|≥1 Implies that -1≤ z1-1 ≥1 Implies that 0≤ z1 ≥ 2 Implies that z1€ [0,2] Similarly,. z2€ [0,4] and. z3€ [0,6] Therefore, for greatest value of |z1|+|z2|+ |z3| ; z1, z2 and z3 must be greatest. Implies that |z1 +z2+z3| =|z1|+|z2| + |z3| =|2| +|4| +|6| =12
My answer x= -1/2then y= √3/2,x=-1/2 then y=-√3/2, x=0 then y=0 , X= 1 then y= 0 sir iske solution 4 aare he hai btw thank you sir appne hum a re liye 3 baar lecture record kara thanku sir ❤️
Last question answer is, Let Z= x+iy , Z bar = x-iy ,. Therefore , (x+iy)²= (x-iy) x²+y²i²+2xyi=x-iy By comparing the real part , X²-x=0 Therefore x=0 and x=1.... By comparing the imaginary part,, 2xyi+iy=0 (2x+1)y = 0 X= -1/2 and y = 0 Therefore possible solution are 1. x=0 and y=0 2. x=1 and y=0 3. x=-1/2 and y=0
z2=z Taking modulus both sides we get ∣z∣2=∣z∣{∣z∣=∣z∣} ∣z∣2=∣z∣ So,∣z∣=0,1 case(1),∣z∣=0,z=0 case(2),∣z∣=1, Then the equation is z2=z=z−1 z3=1 This gives the remaining 3 solution. Hence, the no. of solution of equationz2=z is 4 solution
sir answer for homework, let z=a+ib given z^2=z conjugate then, (a+ib)^2=a-ib (a^2-b^2)+2abi=a-ib so,compaering real and imaginary parts we get a^2-b^2=a eq 1 2ab=-b eq 2 from eq 2 a=-1/2 putting value of 'a' in eq 1 b=+-Root3/2 hence z has 3 possible value in (a+ib) form. sir my name is om kumar singh . from bihar sir comment karke please bata dena ki kya jo mera method sahi hai , i am in class 11 th
Z sq = z bar Modulus both sides Mod (z sq) = mod (z bar) Mod (z sq)= mod (z) (Mod z)sq = mod(z) Mod z = 1 Z = +-1 Sir Aisa kar sakte hain?? Sir waise aap bahut acchha padhate hain... ekdum भन्नाट...
Hey I want to ask can I start my whole syllabus from this channel . Because I didn't study anything . Are they teaching good for chemistry and physics also . Can I get good marks
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Jitne bhi bhaii Argand plane and Polar Representation ki study ke liye ye video dekh rhe hain to unke liye video sidha 22 minutes ke baad start hai. So don't waste your time before.
Sir literally I am watching this video in 2023 and I want to say that this video is very good for students . Those who are watching this in 2023 let me know by giving like to this comment.
Thank you so much sir. Your classes are helping so much where I could learn clearly. Your classes are attractive through your way of teaching and more understandable.
✌✌✌✌✌✌✌✌✌✌✌👉👉👉👉👉👉 Solution of last question is- z^2=zbar (a+ib)^2=a+in a^2-b^2+2abi=a-ib Case-1 ,2abi=-bi a=-1/2 Case-2. ,a^2-b^2=a ......1. now put the value of a in equation-1 Then we get, (1/2)^2+1/2=b^2 1/4+1/2=b^2 b=+(√3/2)and-(√3/2) Then Z= (-1/2+√3/2i )and( -1/2-√3/2) Now the number of solutions is 2
Homework answer is Z is purely real complex number Z = Z bar Solution Z^2 =Z so, (|Z|/Z)^2=Z bar and |Z|^2=Z bar. Z bar and Z. Z bar = Z bar . Z bar so the answer is Z = Z bar and Z is purely real complex number
salute to such a great teacher who recorded lecture 3 times only for us...
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4
True😢
How 3 times
@@MayankisG lecture pura dekha?
@@AyaanKhan-iy4vh no😂
Best Cinematography award goes to Aman Sir 😊😊
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4
How much can i score in school exams with the ncert
@@ravinderpoonia72 school exam me to bht achha krlo ge ncert se entrance me try bhi mt krna bs😂🤣
No no no Oscar a
The solution of homework question is
Let Z=a + ib
So Z conjugate = a-bi
Now Z²=Z conjugate
a² + b²i² + 2aib = a-bi
a² - b² +2aib = a-bi
Now by comparing the real part by real and imaginary part by imaginary
a = a² - b²
b = -2aib
a = -1/2
Putting this in a = a² - b² we get
b = +(√3/2), (-√3/2)
So complex number Z=(-1/2) +(√3/2) i and Z = (-1/2) -(√3/2) i
Correct
0 and 1 will also be its solutions..... all real numbers are also counted under complex numbers
correct
Right 👍 ji
Bro 2 years have gone wha are you doing now nikla jee
Salute sir aapne yah lec 3 baar record kiya sirf hamare liye sir aap logo ki wajah se itna achcha quality education free milta hai aur sir yah aap logo ki mehnat hamare result mein chamkegi
Very hardworking persons in physics wallah team...
If you are preparing JEE/NEET you must watch this video🔥🔥
👇👇👇👇👇
ua-cam.com/video/HebDGCMWPPI/v-deo.html
Sir samajh nahi aata hai😢
sir aap ki mahenat hamare results mein chamke gi.
Sahi bhai
Meri 11th waste ho chuke... Bt 112 m i will be the topper..... Myself jitesh.... Dekh lena i will be the topper.....
😂 😂😂 good joke
@@jiteshydv31 good joke 😂😂😂😂😂😂
You recorded this lecture 3 times only for us.
Thank u Aman sir
36:49
Answer
Omega, 0 and 1
Why because if we consider the Z as a+ib, then we have got a imaginary value i.e
(-1+i√3) devided by 2 = omega
And the other values should 0 and 1
By putting we got
Given That:- z^2 =zconjugate
To Find- Possible Solution Of Z
Soln -
(Step1):- write (Z^2 = Z conjugate) in modulus.
:- |Z|^2 = |Z conjugate|
..by Modulus Property ( |Z Conjugate| =|Z|)
therefore,
:- |Z|^2=|Z| ..(equation)
(Step2)by Observing the above equation
| Z | = 0&1
(Case1)-Put |Z|=0
then Z=0 ..Ans.1
(Case2)- Put |Z| =1
then (Z^2)=(Z conjugate) = (Z^-1)..ans.2
Z^3 =1
so,Z =1...ans.3
Therefore, this gives remaining 3 Solution .Hence |Z|^2 =|Z Conjugate| have over all 4 Solutions of 'Z'
❤️PHYSICSWALLAH
❤️PACE
❤️AMANSIR
Aman sir ,, Alakh sir ,, Amit sir ,, 😂😂. 🇮🇳🇮🇳 Bhaut Bhari combination
Rohit sir???
Oh sorry but I taken those whose name start with A
Hnn brother
Rohit sir too!!!!!!!!
Hmm aur Aman dhattarval or arvind Arora bhi sub a naam ke hai sub ek se ek teacher hai ekdum mast
We can write it as A^5😁
All possible solution of z square =conjugate of z is .
1; if y=0 then x=0
2;if y=0 then x=1
3;if x=-1/2 then y=+√3/2
4;if x=-1/2 then y=-√3/2.
Sir please check my answer.
It is right or wrong.
36:25 z=(1+0i)
Z=(0+iy)
Z=(-1/2+ROOT 3/2i)
Z=(-1/2- root3/2i)
Ye answer kysa aaya aapko
@@jayalakshmi2133 take z=×+iy and conjugate z as x-iy then put z^2=conj.z and solve
Bhai please photo bhej de
I salute 🙏🙏🙏🙏🙏 to these teachers who work so hard to prepare us for JEE and then we have school teachers , good for nothing....
Sir i really respect the hard work and determination you put in your lectures. I am highly thankful to you sir. Thank you so much Alakh sir for finding such gems.
Sir you cannot imagine what is the value of your teaching in our life. You teach mathematics like never before. We all respect you and your hardworking towards us .
If you are preparing JEE/NEET you must watch this video🔥🔥
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How was your exam
Exam diye the if yes then kaisa gaya exam 🤔
Sir I had written solution of questions
(Z^2 = Z conjugate) ..... possible solu
But I will again write the solution
0,1, -1/2 + root 3/2i , -1/2 - root 3/ 2i.
Solve these equation ; ; ; ; ; ; ;
x^2 - y^2 = x...............................
x^2 + y^2 = x...............................
And this property /Z/^2 = Z. Z conjugate. So
0:00 Intro
4:14 HW Qs Discussion
10:14 Modulus Properties
12:33 Proof of Property 6
32:45 Question
36:15 Outro
Thank you so much bro
Z= 0+0i
Z= 1+0i
Z= -1/2 +(root3/2)i
These are the possible complex no. That satisfy the given condition in the last H.W. Question i think.
*salute to your efforts sir..aap hamare liye itni mehanat karte ho love u sir.*
The possible values of Z= 0 , 1 , w(omega) , w^2
w= (-1+root 3 i ) / 2
w^2 = (-1-root3 i ) / 2
Bhai इसका मतलब क्या है?
In which standerd are you readind
*sir* *homework* *question* *answer* *is*
Z=1
Solution
Given Z^2=Z'
Multiplying both sides by Z
(Z)(Z^2)=(Z)(Z')
Now R.H.S will equal to |Z|^2
So
(Z)(Z^2)=|Z|^2
Z= 1
|bhannat answer|
Good boi
Homework: 4solutions
(0+i0), (1+i0), (-1/2+i√3/2),(-1/2-i√3/2)
Putting z=x+iy in the equation and then equating Re(z) and Im (z) we get 2 values of x=-1/2 and 0 and 3 values of y=±√3/2,0
bro bilkul shi hemera bhi yehi aya . pata ni sablo properties learning ke bare me comment kar rahe he parkoi ans ni comment karta. bas match karana tha . thanks
@@niharsanoria2050 For your kind information, total no. of solution is 6
Solution other than yours are 0 - 0i and 1 - 0i
@@aliceverma8065 can you explain the solution pls
ANS TO HOMEWORK QUESTION
Z = ( 0) , (1) , (-1/2 + ROOT3 iota / 2) , (-1/2 - ROOT3 iota / 2)
MY NAME IS MANTHAN . JAY HIND JAY BHARAT
How
Ya my too
@@yearsago-wz3jo assume z equals to a+bi,and then conjugate is a-bi z ,squaring the value of z and put it equal to value of conjugate then compare them u will find a and b.
22:39 properties memorised....bhannat tarike se🙏😎
Thanks ❣ 😊 for giving ur special time to us...3 times recorded same lecture....one like 4 sirr😘😘
Sir,
Last me question ka answer 4 solution hoga.
Z² =z bar {|z bar|=|z|}
Z²=|z|
|Z|=0,1
Case1: |z|=0,z=0
Case2: |z|=1
Z²=zbar =z-¹
Z³= 1
So, total possible solutions = 4
Mereko tera wala Shi nhi lg raha hai
Let's see sahi hota hai ya nahi
I was watching this video on PW App. But after listening, the story of recording lectures 3 times, I come here for give this video a huge like...
And a lot of respect to AMAN SIR.
Bhannat Maths Teacher.
Answer to the question u gave in H.W
Q. z^2=conjugate of z
Let z=a+bi
Then. z^2=a^2+2abi-b^2
Conjugate of z=a-bi
A.T.Q
a^2+2abi-b^2 =a - bi
(a^2-b^2)+(2ab+b)I=a+0i
Equating real and imaginary part,
a^2-b^2=a ..1..
2ab+b=0
a=-1/2
Putting value of a in ..1.. ,we get
b^2=3/4
b=+√3/2 or -√3/2;
z=-1/2+√3/2i
z=-1/2-√3/2i
Values of z :
z= -1/2 +√3/2i and -1/2-√3/2i
And the Oscars for best cinematography goes to Mr. Bhannat Aman sir
If you are preparing JEE/NEET you must watch this video🔥🔥
👇👇👇👇👇
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let z = x + iy then z(conjugate) = x - iy and z^2 = x^2 - y ^2 +2ixy comparing we get,
x = x^2 - y^2 and y = -2xy
if y = 0 , then x = 0,1
if y not equal to zero then x = -1/2 solving for y we get y = [+ or - root(3) / 2 ]
Bro/Bhen mere usme me x ki final value -1/2 aur y ki final values underroot +*-(3/2 ) aa Rahi hai
The guy who is speaking in starting of the lecture (video) has Very Sweet voice 🤗
H.W. Solution-z=0,1,-1/2+√ 3/2 i,-1/2-√ 3/2 i
Just take modulus on both sides to get |z|=0,1 then multiply both sides by z to get z³=1 hence other solution are root of unity (found 1 and solutions of x²+x+1=0 by quadratic formula)
Here instead of x I mean z
Really sir, you are very great.
Hamaare liye ap kitna hard work karte ho.
We promise, hum apka mehnat ko bekar hone nenhi denge.
We will also do hard work with you to be successful.
Thank you❤ sir🥰
❤❤❤❤❤❤❤❤❤
Solution of last question is- z^2=zbar
(a+ib)^2=a+in
a^2-b^2+2abi=a-ib
Case-1 ,2abi=-bi
a=-1/2
Case-2. ,a^2-b^2=a ......1. now put the value of a in equation-1
Then we get, (1/2)^2+1/2=b^2
1/4+1/2=b^2
b=+(√3/2)and-(√3/2)
Then Z= (-1/2+√3/2i )and( -1/2-√3/2)
Now the number of solutions is 2
Sir you are a valuable and priceless gift for us.
the solution of this question 32:56
|Z1-1|
Answer of homework question
z=[-1+3^(1/2)i]÷2
z= [-1-3^(1/2)]÷2
Are all possible solutions
1 and 0 should also be there
Homework question:
Z^2 = Z (bar)
|Z^2|=|Z| {|Z(bar)| = |Z|}
On solving we'll get two cases.
Case 1:- |Z|=0 then Z=0
Case 2:- |Z|=1 then
Z^2 = (|Z|^2)/Z {Z(bar) = (|Z|^2/Z)}
So, Z^3 = 1
And cuberoot of Z will give cube roots of unity that are 1, omega, omega^2.
So there can be 4 values of Z in equation Z^2 = Z(bar) . They are 0, 1, omega and omega^2.
Thank you!!😊
If aman sir teaches us mathematics in class 12 in UA-cam, then our maths will become purely bhannat !
👇😀😀😀
Correct😂
Yeh
Z=a+bi =Z^2=a^2-b^2+2abi
Zconjugate=a-bi
compare both
a=a^2-b^2
b=2abi
to solve this
we get
a=-1/2
and
b=-underroot 3/2
best cinematography award goes to Aman sir😀
Thank you sir for recording it 3 times even though we are not paying u
How 3 times? Only 2 times
Khan sir Research centre
Physics walla- Alakh sir
Super 30 - Anand kumar
Edumantra
If combine 🔥🇮🇳🔥🇮🇳🔥🇮🇳🔥🇮🇳
🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
Aapka mehnath kabhi bee vruda nahi homge sir
Thanks for your dedication on us sir
36:17
Sir we can assume z=a+bi
So we will get
a²-b²+2abi=a-bi
So 2ab=-b.......equation 1
a=-1/2(for b≠0)
Now
a²-b²=a....... equation 2
Put a=-1/2
We will get b= ±√3/4
So
z=-1/2+√3/2
z=-1/2-√3/2
Sir these both complex numbers are also cube root of unity represented as w and w² respectively where w²=w'
Getting back to the question😜
Sir two more answers are possible.....
From equation 1
For b=0
Equation 2
a²-0²=a
a=0,+1
z=0+0i
z=1+0i
So finally there are 4 possible values of z
Sir my name is Shashank
Home work question answer is:
Z=1 ans
Yo yo honey singh 😂😂
Same here
Same here
HW question answer
4 possible value 1 , 0, -1/2+rootunder3/4i , -1/2-rootunder3/4i
explanation:-
(a+bi)^2 = a- bi
a^2 + b^2 +2abi = a - bi ...........(1)
compare imaginary part
2ab = -b
b+ 2ab=0
b(2a+1)=0
so b=0 or
a=-1/2
now put the above value in real part of (1)
you will get two value corresponding above a and b each
so finally you will get 4 possible value of z
Thank you sir
From biology student🙏🙏🙏❤❤ harmre liye itna mehnat karne ke liye😍👍🙏❤
is online classes ke duniya meh aap hi hamare sahara(help) and inspiration ho sir ji hats off to u
33:00
Ans. 12 (*Different Solution*)
Solution: For max value,
|z1 - 1| = 1 ---> z1 = 2
|z2 - 2| = 2 ---> z2 = 4
|z3 - 3| = 3 ---> z3 = 6
Max value of |z1 + z2 + z3| = 12
Same as me bri
Given:-
z2=z
to find the no. of solution of z2=z
Solution:-
Given equation,
z2=z
Taking modulus both sides we get
∣z∣2=∣z∣{∣z∣=∣z∣}
∣z∣2=∣z∣
So,∣z∣=0,1
case(1),∣z∣=0,z=0
case(2),∣z∣=1,
Then the equation is
z2=z=z−1
z3=1
Sir homework ques answer
-1/2 ± whole root -3/4
Z has 2 possible solution
Z has 0 real solutions
How z has 2 solution pls pls explain
Mere ko bhi aisa lagta h par koi sol. Bata dena correctly
@@piyushraj0123 doubtnut.com/question-answer/solve-the-equation-z2-barz-where-z-is-a-complex-number-51237936
check this
mera bhi same ans aaya hai bro (me hu kafi aage bas ye mera question reh gaya tha to socha aaj ek bar aur try karu and i pulled it off)
@@godgamer2258 now got it happy?😊
| Z1-Z2 |
Last Q ans: number of solution is 3
That is =1,0,-1.🙂
How?
Yes mera bhi yahi aaya
-1 is incorrect only 0,1
Solution of last question is- z^2=zbar
(a+ib)^2=a+in
a^2-b^2+2abi=a-ib
Case-1 ,2abi=-bi
a=-1/2
Case-2. ,a^2-b^2=a ......1. now put the value of a in equation-1
Then we get, (1/2)^2+1/2=b^2
1/4+1/2=b^2
b=+(√3/2)and-(√3/2)
Then Z= (-1/2+√3/2i )and( -1/2-√3/2)
Now the number of solutions is 2 ✌✌✌✌✌✌✌✌✌✌✌✌
Properties bhannat tareke se yaad kar liye 👌
Last question answer:
z²=z bar
taking modulus on both sides,
|z²|=|z bar|
Now, |z|=|z bar|
=>|z²|=|z|
=>|z|=0,1
|z|=0 is possible iff z=0
|z|=1
Taking square on both sides,
z×z bar=1
=>z bar=1/z
Also, z²=z bar
=>z²=1/z
=>z³=1
z³-1=0
We can see one solution is z=1
=>(z-1)(z²+z+1)=0
Solving, we get
z=1,(-1/2)+-(√3 iota/2)
=>there are 4 solutions:
z=0, 1, (-1/2)+(√3 iota/2), (-1/2)-(√3 iota/2)
For bieng max value z1 should be equal to 1 or otherwise it will have smaller one then:-
Z1-1=1
Z1=2
Same with z2 and z3 so:-
Z2=4
Z3=6
And adding all will sum up 12 bhannat😀
Soon 5 million family ..
Who knows that, it has uploaded again, after deleted video....
Friend it's uploaded again,I watched old video so I need to watch it again (is this video has something extra? )
@@aarishansari454 no
@@diedforu thanks I'm just going to watch it in few minutes after my physics lecture
Thanks yaar time save karne ke liye🙂🙂
4:53 sir maine bheja tha solution ke sath answer... ab mai check karunga
22:27 sare properties ekdum bhannat yadd hoo gayi hai
Salute h sir aapko hamare liye kitna hard work kr rhe ho aap log ham bhi sir aapko competition fod ke dekhiyenge🤗🤗
Hats off sir....
Hmare liye 1 hr k lecture 3 times record kiya ....
That's called inspiration.......
36:40 my answer is coming 1
But I think that all the real number should satisfy the equation
And also the equation given in question will be purely real
my answer is 0 and 1
Bhai mere 4 solutions arahey hain
No bro how can all real number, lhs will be a² and rhs will be just a
For that last question ....
So for greatest value
|z1+z2+z3| must be
=|z1|+|z2|+|z3|
Now a/c to ques.
|z1-1|≥1
Implies that -1≤ z1-1 ≥1
Implies that 0≤ z1 ≥ 2
Implies that z1€ [0,2]
Similarly,. z2€ [0,4]
and. z3€ [0,6]
Therefore, for greatest value of
|z1|+|z2|+ |z3| ;
z1, z2 and z3 must be greatest.
Implies that
|z1 +z2+z3| =|z1|+|z2| + |z3|
=|2| +|4| +|6|
=12
My answer x= -1/2then y= √3/2,x=-1/2 then y=-√3/2, x=0 then y=0 , X= 1 then y= 0 sir iske solution 4 aare he hai btw thank you sir appne hum a re liye 3 baar lecture record kara thanku sir ❤️
Ans of HW QUESTION
there r 2 values of Z
1...-1/2+root3/3 i
2...-1/2-root3/2i
22:27sir yaad hogye saari properties
Last question answer is, Let Z= x+iy , Z bar = x-iy ,. Therefore , (x+iy)²= (x-iy)
x²+y²i²+2xyi=x-iy
By comparing the real part ,
X²-x=0
Therefore x=0 and x=1....
By comparing the imaginary part,,
2xyi+iy=0
(2x+1)y = 0
X= -1/2 and y = 0
Therefore possible solution are
1. x=0 and y=0
2. x=1 and y=0
3. x=-1/2 and y=0
36:21 ans of the homework question is z=(-1/2 ) +- { (root 3) / 2 }
sorry ajeeb tareeke se likha he vo comp se type kar raha tha . :-)
Hw ka answer.... - 1/2+root3/2i , -1/2-root3/2i
May God bless 🙏 Aman sir and his family .
z2=z
Taking modulus both sides we get
∣z∣2=∣z∣{∣z∣=∣z∣}
∣z∣2=∣z∣
So,∣z∣=0,1
case(1),∣z∣=0,z=0
case(2),∣z∣=1,
Then the equation is
z2=z=z−1
z3=1
This gives the remaining 3 solution.
Hence, the no. of solution of equationz2=z is 4 solution
Day by Day entry becomes just wow 😍 no words 🤐 u are the real hero sir salute
36:39 (x, y) ={(-1\2, √3\2), (-1\2, -√3\2) } ans hai sir😇😇
Absolutely correct
What is your method of doing
I did by assuming z= x + iy
@@meenakshigupta3221 Dekh lo bro
sir answer for homework,
let z=a+ib
given z^2=z conjugate
then, (a+ib)^2=a-ib
(a^2-b^2)+2abi=a-ib
so,compaering real and imaginary parts we get
a^2-b^2=a eq 1
2ab=-b eq 2
from eq 2
a=-1/2
putting value of 'a' in eq 1
b=+-Root3/2
hence z has 3 possible value in (a+ib) form.
sir my name is om kumar singh .
from bihar
sir comment karke please bata dena ki kya jo mera method sahi hai , i am in class 11 th
Bhai congucate kaha hoga square hai na ?
@@bloomapnavidyalaya-30M hai question dekho last wallah 36:17
@@nobeldude dost , z squqre is not equal to z bar
Jyada difference ho jayega
@@bloomapnavidyalaya-30M bhi question me given hai usi se kiya hai
@@bloomapnavidyalaya-30M bhi z bar ka matlb z ka conjugate hi hota hai na
Z sq = z bar
Modulus both sides
Mod (z sq) = mod (z bar)
Mod (z sq)= mod (z)
(Mod z)sq = mod(z)
Mod z = 1
Z = +-1
Sir Aisa kar sakte hain??
Sir waise aap bahut acchha padhate hain... ekdum भन्नाट...
Very Excited to see this lecture.
ua-cam.com/video/Eu17EM5MApI/v-deo.html
Hey I want to ask can I start my whole syllabus from this channel . Because I didn't study anything . Are they teaching good for chemistry and physics also . Can I get good marks
Hey guys this comment is for you just visit this video only if you are a serious pace batch student and also see yourself successful in future otherwise you may skip this video .. ua-cam.com/video/PssDEPkn3JQ/v-deo.html ua-cam.com/video/PssDEPkn3JQ/v-deo.html
4
@@arshpreet5791 100% u must start with this if u have enough time and dedication
Answer of hw problem is:-
Z=0,1,±root3/2
Plzzz give us sigma lecture I am rally facing problem on this concept 🙏🙏🙏🙏. Who else want hit like
Jitne bhi bhaii Argand plane and Polar Representation ki study ke liye ye video dekh rhe hain to unke liye video sidha 22 minutes ke baad start hai. So don't waste your time before.
Sir next chapter PERMUTATIONS AND COMBINATIONS
Next chapter is Quadratic Equation and Linear inequality 😒ncert nahi Kiya?
Ncert pagal h
@@Ken_77 ya quadratic is too imp for jee
Mera bhi Permutations ka hai Par pehle Linear Equalities Hoga..Vo Zyada Important Hai. Bhai Aap Baniye Ho Kya? Me Bhi Baniya Hu.
last question ans.......let z=x+iy than solutions x=1/2 and y=+ -1/root2
22:50 all properties learnt.
If you are preparing JEE/NEET you must watch this video🔥🔥
👇👇👇👇👇
ua-cam.com/video/HebDGCMWPPI/v-deo.html
Achha ladka aur achha intelligent boy mere jaise nahi milta 😂😂
answer to the homework question is that possublecomplex number are (-1/2 +root3/2 i),(-1/2-root3/2 i),(o+oi),(1+0i)
Sir literally I am watching this video in 2023 and I want to say that this video is very good for students . Those who are watching this in 2023 let me know by giving like to this comment.
bhai abhi 2022 hi hai 14 dec
@@kartikeypal1175 yaa right 😂
@@kartikeypal1175 🚩🚩bhai 2079 chal raha
Answer of the last question is -1/2 + (+_ √3/2 )i
How?
मेरा तो + - 1 आया है
Thank you so much sir. Your classes are helping so much where I could learn clearly. Your classes are attractive through your way of teaching and more understandable.
✌✌✌✌✌✌✌✌✌✌✌👉👉👉👉👉👉 Solution of last question is- z^2=zbar
(a+ib)^2=a+in
a^2-b^2+2abi=a-ib
Case-1 ,2abi=-bi
a=-1/2
Case-2. ,a^2-b^2=a ......1. now put the value of a in equation-1
Then we get, (1/2)^2+1/2=b^2
1/4+1/2=b^2
b=+(√3/2)and-(√3/2)
Then Z= (-1/2+√3/2i )and( -1/2-√3/2)
Now the number of solutions is 2
1 ch 9 detailed videos 👏👏👏👏 you are such a great teacher sir..
Homework answer is Z is purely real complex number Z = Z bar Solution Z^2 =Z so, (|Z|/Z)^2=Z bar and |Z|^2=Z bar. Z bar and Z. Z bar = Z bar . Z bar so the answer is Z = Z bar and Z is purely real complex number
Sir aap humare liye itna mehnat kar rahe hai. Thank you sir
Ha sir yaad ho gai sari properties
Sir we appreciate your efforts for making this video 🔥🔥🎉🎉
last homework question
z²=z'
z=x+0i
z'=x-0i
z²=x²+0i² = x²
x²=x
x= 1,0
I'm in 10th 😜i hope my answer is correct
Yes 👍
But concept is wrong
But now you are in 11
My classmate
33:03
Sir maine aise kiya hai
drive.google.com/file/d/1-LybG6Y-LYJtp70S5I7quw4W0BBsw9wL/view?usp=drivesdk
Good 👍
Bhai tum kaise ye link lagaye ho
@@shubhampandey4633 Google drive pe upload karke link ko copy kar lena aur yahi link clipboard mein niklega toh de dena
@reemlin konwar thanks bhai
@@shubhampandey4633 welcome
36:19 Sir, there are two solutions, one is z=-1/2 + root(3)/2 * i, and the other is z = -1/2-root(3)/2*i
Suppose Z is equals to X + iy after solving further I got 3 solutions Z=-1/2+_root3/2i and 0
Nhi bro answer 0,+1,-1 hoga
@@Samarth-bg7zt Answer is -1/2+root3i/2
Thanx sir lec 3 upload karme ke liye
Sir thoda sa mistake ho gaya aapse . z1. z2= (a1a2-b1b2) +(a1b2+b1a2) hoga🙂