1. ## complex variable

im struggling with what seems to be an easy question
it asks to find [COLOR=rgb(0, 0, 0)]all solutions to the following equations
- (e^2z)-2i=0
-(z-i)^5=1
are two im sure there not hard because he has given us ten to do but just don't see how to do them or is it as simple as jsut breaking them down and putting z=x+iy
[/COLOR]

2. Just to be clear, is the first problem $e^{2z}-2i=0~?$

3. ya think if i could figure out one id be able to go with the rest of them

4. That is the same as $e^{2z}=2i$ so $2z=\log(2i)$.
Do you understand the complex logarithm?

5. not really ill look it up

6. in these questions am i just looking for values of z yes??

7. if its sin(z)+2i=0 any idea how id get rid of the sin? sorry last question this stuff is hard!!

8. $\sin(z)=\sin(x)\cosh(y)+i~\cos(x)\sinh(y)$