How can polar representation help solve complex, exponential problems?

[nickolas2730]

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

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

But i don't understand how does these 2 steps work
<=>e^2iz=-1
<=>2iz=πi +2kπi

how can a exp function suddenly become πi +2kπi

Thanks

 

[nickolas2730]

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

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

 

[micromass]

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]

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

Yes, this is what you use.

But how can i know in exam -1 = e^πi
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]

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

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

But i don't understand how does these 2 steps work
<=>e^2iz=-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: $\displaystyle -1=e^{(\pi i)}e^{2\pi k i}\,.$

 

[nickolas2730]

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 e^πi=-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]

thanks, i know e^πi=-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]

thanks, i know e^πi=-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 let's 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.