Complex numbers : Finding Square roots of : ExamSolutions
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- Опубліковано 21 сер 2012
- Tutorial on how to find the square roots of complex numbers.
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Because i equals the root of -1 and so when you square this you get -1... -1 multiplied by b squared gives - b squared. Take a look at my early videos on complex numbers to see these basic results for squaring, cubing etc for values of i.
Thank you that was really helpful. Excellently explained, thanks again.
I'm not sure. I just wanted an explanation on how to factorise those quartic equations that you did in this video
Use u substitution. Substitute b^2 for u and it becomes a quadratic. Solve for u. Once solved for u, replace u again with b^2 and then solve for b.
@@AliShaikh_744 Thank you brother 😊❤️, I really appreciate it, I now understand
@@emmydee408 Glad I was able to help😁
Thank you very much!
you are a saviour 🙏🙏🙏
Very helpful.
i like the way u interpret the question
Thank you!
Do you have a video up on how to factorise quartic equations?
i needed this
Thanks
How can I pay you money?
you can now donate to him! please consider if possible though it's completely understandable if you cant due to the pandemic :)
Thank you man may god bless u😊
Thanks mate
No problem
Thanks a lot
Can't believe this was uploaded when i was 10 years old
b is a real number. For example 2i is an imaginary number but the 2 is real
Quality video. Thanks for the help.
Cheers
+faiza anwar At 6.44 I said that b was a real number. It is not an imaginary number. 3i is imaginary but the 3 is real. So bi is imaginary but b is a real number. I hope that makes sense.
You sir are a legend thank you so much
Best wishes - thanks for your support.
Do you mean ones with complex roots?
but if you substitute B= -5, +5 into the first equation (a^2-b^2= -21) you would only get a=2 right?
Super
... if I may ask... why isn't b = +/- 2i one of the solutions ?
Sir, may I know at the beginning part, how did you make that ...+b squared i squared reduced to -b squared?
I know it's been 9 years, but seriously? That was the most difficult part to understand?
It's because it's (bi)² so b²i² i²=-1 so it's b²×-1 so -b²
There are 2 types of people lmao
e^ipi = -1, i^2ipi = 1, 4(e^2ipi) = 4
nice
😭😭😭you saved me
Oh okay, and now i noticed i made a typo in the second statement :O that i should be an e....
why does b have to be a real value 6:40?
It will yield the same result by considering only b real. The a and b will only be swapped
If b is not real then it would make a+bi into just a+b (as i x i is real) and so it won't make it correct statement
Safe Rehman ah ok
if i want to find the square root of 1+root of 2 for example can i use a similar method?
Yes
Sorry, my brother logged into my e-mail account and replied to this. Rest assured, I understood it. Well played sir.
why solution for square root of y must be real number ? 6:35 and 11:30
If b isn't real and has an i component, the final answer would be a+bi^2. Which is equal to a-b, and there would be no imaginery parts in the final answer, thus it would not be a complex number[
@@junaidwali8938 you're about 3 years and 6 months late
... time travel@@radheypatel9622
@@radheypatel9622 better late than never😅
gud
Sugene Kim approved
-(i^2) st.
but if you use b=12/a, the answer is ±(4+3i) instead of ±(4-3i)
does it matter?
:) thanks
b = -12/a not 12/a
Haha it seems so! I didn't bother to look, I read the first comment anyway.
what if there was a "z" in the complex equation ?
Z is not part of a complex equation z IS a complex equation
*first ;)
Why did the sign not change when the 21b^2 switched sides
did you ever figure this out haha
Because it didn't switch sides, he added b⁴ and and subtracted 100 to make it 0=b⁴-21b²-100 but just wrote it the other way round as b⁴-21b²-100
???
f
Or we can use polar forms