Question

Express log(2x^2-2y^2+1 + i4xy) in the format u +iv where i is the imaginary number

Answer 1

and \(x, y \) are real?

Answer 2

yes

Answer 3

hmmm

Answer 4

\(\log(z)=\ln(|z|) + i \arg(z)\)

Answer 5

and of course log is a multi valued function, because \(\arg(z)\) is not unique

Answer 6

\[z=2x^2-2y^2+1 + i4xy\]
\(re(z)=2x^2-2y^2+1\) and \(im(z)=4xy\) so \(|z|=\sqrt{(2x^2-2y^2+1)^2+(4xy)^2}\) some algebra is needed here

Answer 7

i thought this would turn out to be an nice perfect square or something, but it doesn't
i wonder if i am doing something wrong
it is just what it is

Answer 8

Can you find a branch then of log(2*z^2+1) that is analytic at i

Answer 9

now you are making me think

Answer 10

My next question is to find the derivative of log(2z^2=1) at i. My answer is to use the chain rule and to substitute i with an answer = -4i. Is this the correct way to go?

Answer 11

yes to your second question
the derivative will be \(\frac{1}{2z^2+1}\times 4z\) but i am still thinking about what branch cut you need, and why you cannot do the usual cut on the negative reals

Answer 12

I am waiting, I thing you can help me with this.

Answer 13

i just need to find an example. i think this is straight forward. let me check alfors

Answer 14

damn not there. ok, do you have a similar example to work from?

Answer 15

No, not at this time. I am going to post another complex analysis question as a new question. We can leave this one, maybe you will solve it late. I first would like to no where do you get the nice mathematical notations in your response seeing that these brackets/squares and exponentiation signs confuse me at times. Thank you for assisting on this problem

Answer 16

i am going to keep looking and see if i can find how to do this. i think it is a simple matter of finding the zeros of your inside function and cutting there , but i could be wrong

Answer 17

maybe we can mimic this

Answer 18

wrong one, i meant this one, problem 4

Answer 19

Thank you very much. Would you please check the new problem for me.