Відео

stupid music always slows my thought

Answer = 1-i, (1+i)^4 = -4, (1-i)^4 = -4i, (1+i)^2024 / (1-i)^2023 = (-4)^506*(1-i)/(-4i)^506 = (1-i)

P=given formula P=[(1+i)/(1-i)]^2024(1-i) (1+i)/(1-i)=i i^2024=1 P=1-i

Ma perché scrivete sempre i problemi ditti notevoli? Dovrebbero essere contenti sciuti se fai la n test d'entrata e se non li sai è giusto che stai fuori

[(1 + i)/√2]⁸ = [(1 - i)/√2]⁸ = [(1 + i)/(1 - i)]⁸ = 1 so (1 + i)²⁰²⁴/(1 - i)²⁰²³ = (1 - i)•(1 + i)²⁰²⁴/(1 - i)²⁰²⁴ = (1 - i)•[[(1 + i)/(1 - i)]⁸]²⁵³ = (1 - i)•(1)²⁵³ = 1 - i

(2024i+^2024)/(2023 ➖i^2023)=(2024i^2024/(0+0 ➖ {i^0+i^0 ➖}=(2024i^2024)/(1i^1) (1^1^2^2i^1^2^12)/(1i^1) (1^1^i^2^2^6)/(1i^1) (i^1^12^3)/(1i^1) (i^2^3)/(1i^1) (i^2^3 ) (x ➖ 3ix+2i). I enjoyed avery education mathematics thank you.

Oxford entrance exam question: [(1 + i)²⁰²⁴]/[(1 - i)²⁰²³] =? (1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i, (1 - i)² = 1 - 2i + i² = 1 - 2i - 1 = - 2i [(1 + i)²⁰²⁴]/[(1 - i)²⁰²³] = [(1 + i)²(1 + i)²⁰²²]/[(1 - i)(1 - i)²⁰²²] =[(1 + i)²/(1 - i)]{[(1 + i)/(1 - i)]²⁰²²} = [(1 + i)²/(1 - i)]{[(1 + i)²/(1 - i)²]¹⁰¹¹} = [(2i)/(1 - i)]{[(2i)/(- 2i)]¹⁰¹¹} = [(2i)/(1 - i)][(- 1)¹⁰¹¹] = (- 2i)/(1 - i) = [(- 2i)(1 + i)]/[(1 - i)(1 + i)] = [(- 2i)(1 + i)]/(1 - i²) = [(- 2i)(1 + i)]/2 = - i(1 + i) = - i - i² = 1 - i

풀이가 틀렸다. 답이 1+¿될려면 분모가 1+¿가 되어야 한다.

Nice, but I don't see an advantage in using the "trick" presented here over the direct calculation as follows: x^2 = (sqrt(2)-1)^2 = 2 - 2 sqrt(2) + 1 = 3-2 sqrt(2); x^4 = (x^2)^2 = (3-2 sqrt(2))^2 = 9 + 8 - 12 sqrt(2) = 17-12 sqrt(2); x^6 = x^2*x^4 = (3-2 sqrt(2))*(17-12 sqrt(2))= 99-70 sqrt(2), hence x^12=(x^6)^2 = (99-70 sqrt(2))^2 = 99^2+2*70^2 - 2*70*99 sqrt(2) = 19601-13860 sqrt(2) # Or, even shorter, x^12 = ((x^3)^2)^2.

Oxford entrance exam question: √(- i) =? - i = (- 2i)/2 = (1 - 2i - 1)/2 = (1 - 2i + i²)/2 = [(1 - i)²]/2 = [(1 - i)/√2]² √(- i) = √{[(1 - i)/√2]²} = ± (1 - i)/√2 = ± (√2 - i√2)/2 Final answer: √(- i) = (√2 - i√2)/2 or √(- i) = (- √2 + i√2)/2

98% just like the problem and don't care how many failed.

Elegant! I wanted to multiply out top and bottom completely with a similar pairing trick to what you used but you showed that that would be cumbersome and unnecessary extra work!

If you think of -i in polar form it's a hell of a lot easier.

Good solution 4:39

I used method 2 to calculate this is my mind & I did it in approx . 2 min 🎉

Well, just visualize it. Angles are added. 270/2 = 135 = 90+45. So that’s -1+i right there. Lengths are multiplied so scale by sqrt(2). (-1+i)/sqrt(2). The other solution is just a mirror. I.e. -45 degrees.

I used a third method (basically the method you use to denest radicals for real numbers applied to complex numbers): just set sqrt(3-4i) = sqrt(x)-sqrt(y). When you square both sides you get 3-4i = x+y-2sqrt(xy), so x+y=3 and xy=-4, which gives an easy second degree polynomial to solve: x²-3x-4=0 or (x-4)(x+1)=0. This gives 2 solutions: x=4, y=-1 or x=-1, y=4. Taking the square roots you get sqrt(3-4i)=2-i or i-2.

You lost me around 6:51 . Where did you get x^2 + y^2 =25 You wrote something in the right that is not visible

√ ( 2 ^2 + i.^2 - 2 * 2 * i) = (2 - i), - (2 - i)

Or you could just start from -i = e^i(-pi/2)

Dragged out and thus tediously boring

(x ➖ 1ix+1i).

Once you see the trick to get an x^2 = 1 - 2x relationship - you could just use the binominal theorem to go right to the answer... it's having the insight to get to this point to break the problem down to something that is just straight algebra. Nicely done!

No. The convention is that √(i) denotes the principal value of the square root of i, which is (1+i)/√2. i = exp(πi/2) so √(i) = exp(πi/4) = cos(π/4) + i.sin(π/4) = 1/√2 + i/√2 = (1+i)/√2. The other square root of i is just -(1+i)/√2. There can only be two values for any square root, not the four that you imply. And please quit the ridiculous click-bait. It's not an "Oxford entrance exam question", whatever you imagine that to be.

I used (a^2 + 2ab + b^2)

Although the answer is correct, I saw something wrong as x^12=根號2-1 at 9:44

My question -- this is how I see it: Problem -- a+b=6sqrt(ab) Divide all by b -- a/b + b/b = 6sqrt(ab)/b Reduce -- a/b + 1 = 6sqrt(a/b) All to LHS - a/b -- 6sqrt(a/b) + 1 = 0 Let m = sqrt(a/b) -- m^2 - 6m + 1 = 0 Quadratic -- m = -(-6)/2 +/- sqrt((-6)^2-4(1)(1))/2 Reduce -- m = -3 +/- 2sqrt(2) Equivalence -- sqrt(a/b) = -3 +/- 2sqrt(2) Square both sides => a/b = 17 +/- 12sqrt2 What assumptions (besides that a & b are >0) do I need to make to solve it in this way?

Brill. Method 2 preferred. Thanks

If you consider the complex z-plane the answer is immediately clear : for the 1/3 -power you have to divide the corresponding angle by 3: π/2 * 1/3 = π/6 :hence the answer is cos π/6 +i*sin π/6 =1/2*(√3 + i). In your video you give the solution of z^3 = i, but the third root is uniquely defined, as the solution given here.

👍

using r*cis(θ) works well, I suggest, and gives at once sin^2(θ)=cos^2(θ) and 2r*sinθcosθ=1 so r = 1 and θ=m*pi/4 with m in (1,3,5,7)

(x^4 ➖ 256)={x^0+x^0+x^0+x^0x^0 ➖ x^0 ➖ 'x^0 ➖ x^0}=x^4 (x ➖ 4x+4).

Trivial! Draw a picture: sqrt(i) is clearly (1+i)/sqrt(2). It's simple geometry.

97% failed?bro is capping this shit easy

you can do it by trigonometric form and moivre

Tanks mr❤❤🎉

Nice❤🎉

Why make it so complex. It does not take any imagination to see that x=+2 is an answer.

{4x+4x ➖ }+{x+x ➖ }={8x^2+x^2}=8x^4 2^3x^2^2 1^1^1x1^2 1x^2 (x ➖ 2x+1) .

Very good calculation.

The exact result is 1 and I solved it in my head. By the way, I'm not a math genius! You just have to know the power laws and the third binomial formula! (√2-1)^12=(√2-1)^2*(√2-1)^2*(√2-1)^2*(√2-1)^2*(√2-1)^2*(√2-1)^2=1*1*1*1*1*1=1

Ugh that loud annoying music!!!

Stupid music.

Could you please lower the volume of the music?

(3 ➖ 1)^2^6 (3 ➖1)2^2^3 (1 ➖ 1)1^2^3 ()2^3 (x ➖ 3x+2).

(1,4142 -1)^12 0.4142^12 I guess Oxford isn't my cup of tea. 😄😄😄

I would to like this: ((2)^(1/2)-1)^12=x ((2)^(1/2)-1)=x^12 solve 0.4142...=x^12 which have 2 real and 8 complex solution. Choose the positive real solution. Your way is too complicated. You course you need a calculator, but it is way much is easy.

What a complete mess.

If you write i = exp[ i π/2] ,then it follows that √ i = exp[ i π/4]= cos π/4 +i sin π/4 = √2/2*(1+i) . Note : there is only one value for √ i .There would be two values if you solve z^2 = i .

... coś tu spierdoliłeś !!! według ciebie (sqrt(2)-1)^12 to około 0,188, przecież to bzdura !!!!!!!!!