# Complex, exponential problem

Prove that cos z = 0 if and only if z = π/2 + kπ, where k is an integer.

The solution is :
cos z = (eiz+e-iz)/2 = 0
<=>eiz+e-iz)=0
<=>e2iz=-1
<=>2iz=πi +2kπi
<=>z=π/2 + kπ

But i dont understand how does these 2 steps work
<=>e2iz=-1
<=>2iz=πi +2kπi

how can a exp function suddenly become πi +2kπi

Thanks

i know there is an equation :
ez1=ez2 holds if and only if z1=z2 + 2kπi

But how can i know in exam -1 = eπi
is it by try an error

micromass
Staff Emeritus
Homework Helper
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??

micromass
Staff Emeritus
Homework Helper
i know there is an equation :
ez1=ez2 holds if and only if z1=z2 + 2kπi

Yes, this is what you use.

But how can i know in exam -1 = eπi
is it by try an error

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

$$e^{ix}=\cos(x)+i\sin(x)$$

if you put in $x=\pi$ in the above, then it follows.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Prove that cos z = 0 if and only if z = π/2 + kπ, where k is an integer.

The solution is :
cos z = (eiz+e-iz)/2 = 0
<=>eiz+e-iz)=0
<=>e2iz=-1
<=>2iz=πi +2kπi
<=>z=π/2 + kπ

But i dont understand how does these 2 steps work
<=>e2iz=-1
<=>2iz=πi +2kπi

how can a exp function suddenly become πi +2kπi

Thanks
Multiply $e^{iz}+e^{-iz}$ by $e^{iz}\,.$

$\displaystyle e^{iz}\left(e^{iz}+e^{-iz}\right)=e^{2iz}+1$

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

Yes, this is what you use.

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

$$e^{ix}=\cos(x)+i\sin(x)$$

if you put in $x=\pi$ in the above, then it follows.

thanks, i know eπi=-1

but the problem is , when i was in exam i was looking at the -1 , how can i know e?, what about the question is -2 , do i need to try it one by one like π/2,π/3....

micromass
Staff Emeritus
Homework Helper
thanks, i know eπi=-1

but the problem is , when i was in exam i was looking at the -1 , how can i know e?, what about the question is -2 , do i need to try it one by one like π/2,π/3....

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

$$-2=2*(-1)=e^{\log(2)}e^{i\pi}=e^{\log(2)+i\pi}$$

So $\log(2)+i\pi$ 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.

BruceW
Homework Helper
thanks, i know eπi=-1

but the problem is , when i was in exam i was looking at the -1 , how can i know e?, what about the question is -2 , do i need to try it one by one like π/2,π/3....

Polar representation is the key. So lets say in exam, you're given some number like $a+ib$ then this is equal to $r(cos(x)+i \ sin(x))$ which is also equal to $r \ e^{ix}$ 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.