Question

Find a branch of log(2(z^2) +1) that is analytic at i (imaginary number) and compute the derivative at i.

Answer 1

try checking if the given fx is analytic or not ..

Answer 2

*fz

Answer 3

Could you give me more detail please.

Answer 4

let z = x+iy ... simplify it into u(x,y)+i v(x,y) and you know what to do.

Answer 5

ln(2 ((x^2-y^2) + i 2xy)+1) = ln(2(x^2-y^2)+1 + i 4 xy) = ln w = ln |w| + i (arg(w) + 2npi)

Answer 6

Thank you very much.

Answer 7

this is nasty :(((

Answer 8

I am sorry . I tried to give you the best response but the computer is not responding
appropriately. Many thanks.

Answer 9

no probs ... i didn't say you are nasty ... the problem is nasty.

Answer 10

On the problem to find the branch of log (2z^2+1) ending up with
ln(2(x^2-y^2) +1 + i4xy) for the derivative should I only differentiate the real part (ln2(x^2-y^2)+1 and also that for the imaginary part 4xy ?

Answer 11

do you know how to express ln(x+iy) as u + i v?

Answer 12

No I am a new student in complex analysis. Please help. I must still study the concept of residues which is part of the integral questions.

Answer 13

ln(x+iy) = ln sqrt(x^2+y^2) + i arctan(y/z) + i 2 pi

Answer 14

ln(x+iy) = ln sqrt(x^2+y^2) + i arctan(y/z) + i 2 n pi where n=0,1,2 ...

Answer 15

Did you mean arctan (y/x) ?

Answer 16

yep

Answer 17

Thank you very much. This was excellent. I must study and go through all these problems. This is a lonely yourney in Africa with so few people doing this subject. Good night

Answer 18

gotta sleep too ...

Answer 19

best of luck!!