Complex, exponential problem

  • #1
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
 

Answers and Replies

  • #2
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
 
  • #3
22,089
3,296
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??
 
  • #4
22,089
3,296
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:

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

if you put in [itex]x=\pi[/itex] in the above, then it follows.
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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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 [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]
 
  • #6
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.
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....
 
  • #7
22,089
3,296
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 [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.
 
  • #8
BruceW
Homework Helper
3,611
120
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 [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.
 

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