- #1

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Complex Analysis:

Find all solutions of:

(a) e^z = 1

(b) e^z = 1 + i

Find all solutions of:

(a) e^z = 1

(b) e^z = 1 + i

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- 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

Find all solutions of:

(a) e^z = 1

(b) e^z = 1 + i

- #2

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Have you at least made attempts to solve them?

- #3

arildno

Science Advisor

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How fascinating!

We are all desirous to see what you have done so far!

We are all desirous to see what you have done so far!

- #4

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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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Actually y= [itex]\pi K[/itex], since [itex]sin(\pi)= 0[/itex] also. Now, what values can cos y have? And then what is eSo 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!

No, eThe 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...

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