Find solutions of e^z = 1

  • Thread starter racland
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  • #1
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Complex Analysis:
Find all solutions of:
(a) e^z = 1
(b) e^z = 1 + i
 

Answers and Replies

  • #2
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Have you at least made attempts to solve them?
 
  • #3
arildno
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How fascinating!
We are all desirous to see what you have done so far! :smile:
 
  • #4
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So far I got:
e^z = 1
I know e^z = e^x (cos y + i Sin y)
Then,
e^x cos y = 1
e^x sin y = 0
I know that for sin y = 0, y = 2 pi * K, where k is an integer.
After this step I'm stuck!

The second problem: e^z = 1 + i
e^x cos y = 1
e^x sin y = 1
I order for sin y = 1, y = pi/2. Now what...
 
  • #5
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can you express the RHS in polar form?
 
  • #6
HallsofIvy
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So far I got:
e^z = 1
I know e^z = e^x (cos y + i Sin y)
Then,
e^x cos y = 1
e^x sin y = 0
I know that for sin y = 0, y = 2 pi * K, where k is an integer.
After this step I'm stuck!
Actually y= [itex]\pi K[/itex], since [itex]sin(\pi)= 0[/itex] also. Now, what values can cos y have? And then what is ex?


The second problem: e^z = 1 + i
e^x cos y = 1
e^x sin y = 1
I order for sin y = 1, y = pi/2. Now what...
No, exsin y= 1 does NOT tell you that sin y= 1! You might try this: square each equation and add. That will get rid of y so you can find ex (you don't really need to find x itself).
 

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