Question

Give a description or schetch of∶
{z:Im(z ̅^2<1)}
and {z ∶ (z-i) ^2 - (z+i)^2≤1 } as seperate sets
of complex numbers and indicate as open or closed sets

Answer 1

let z = x+iy
z conjugate = x - iy
im(z conjugate) = i 2xy = 2xy

Answer 2

2xy <1 ... how does this look like? ... this is open set. I guess

Answer 3

(z-i) ^2 - (z+i)^2 = -4ixy <= 1

Answer 4

* -4iz <= 1

Answer 5

probably this will give you a circle with radius 1/4 ... and this is closed set.

Answer 6

*i guess

Answer 7

I missed out on the two problems last night.
(1-i)^(i*sqrt{2}) with i=imaginary number ; to prove that all values lie on a straight line
and to find a branch of log(2z^2+1) that is analytic.

Answer 8

http://math.stackexchange.com/questions/189703/does-ii-and-i1-over-e-have-more-than-one-root-in-0-2-pi/189717#comment437924_189717
you will have a single root in [0, 2pi]

Answer 9

that expression equals
http://www.wolframalpha.com/input/?i=++%281-i%29^%28i*sqrt {2}%29+

Answer 10

keep adding 2pi to it ... I don't think it will have any other root in 0, 2pi

Answer 11

as for this problem .. log(2z^2+1)
i think simplification would be the best thing you can do.

Answer 12

\[ \log(Z) = \ln(z) + i arg(z+2 n\pi) \] probably use this formula to do it fase.

Answer 13

it's quite awful to simplify it ... try following the yesterday's procedure.

Answer 14

by letting z=x+iy << turn that expression into u(x,y) + i v(x,y)

Answer 15

and about your earlier Q ... i think i kinda remembered few things

http://openstudy.com/users/hennie#/updates/50430e6de4b000724d461e23

Answer 16

after finding critical points, put those in place of coefficients ... and follow the usual procedure. find Eigen values ... once you know them, you know their nature.

Answer 17

Thank you. This helped a lot.

Answer 18

i hate non linear systems ...