Complex, exponential problem

It's hard to explain why those things hold because I don't know what I can use and what I cannot use.

How did you define the exponential?? What properties have you already seen about the exponential??
Is there a book you're reading where this appears in?? Lecture notes??

 

Yes, this is what you use.

Knowing this depends on what you know about the exponential. I can define for real x:

[tex]e^{ix}=\cos(x)+i\sin(x)[/tex]

if you put in [itex]x=\pi[/itex] in the above, then it follows.

 

Multiply [itex]e^{iz}+e^{-iz}[/itex] by [itex]e^{iz}\,.[/itex]

[itex]\displaystyle e^{iz}\left(e^{iz}+e^{-iz}\right)=e^{2iz}+1[/itex]

For the second question: [itex]\displaystyle -1=e^{(\pi i)}e^{2\pi k i}\,.[/itex]

 

So you want to know what x solves [itex]e^{z}=-2[/itex]? Without going into complex logarithms, this can be solved as

[tex]-2=2*(-1)=e^{\log(2)}e^{i\pi}=e^{\log(2)+i\pi}[/tex]

So [itex]\log(2)+i\pi[/itex] will be a solution (but it will not be the only one).

This can of course also be found by complex logarithms, but you shouldn't use those until you've seen it.

 

Polar representation is the key. So lets say in exam, you're given some number like [itex]a+ib[/itex] then this is equal to [itex]r(cos(x)+i \ sin(x))[/itex] which is also equal to [itex]r \ e^{ix}[/itex] So this is how you convert from one form to the other.

So your example of -2 clearly has the same x as for -1, but the only difference is that r is twice as great.