- #1
racland
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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
What are the solutions for e^z = 1 and e^z = 1 + i?
- Thread starter racland
- Start date
In summary, the conversation discusses finding all solutions for two equations involving complex analysis. The first equation is e^z = 1 and the second is e^z = 1 + i. The conversation also includes attempts at solving the equations and suggests using polar form and squaring equations to find solutions.
- #2
neutrino
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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!
We are all desirous to see what you have done so far!
- #4
racland
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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...
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
JonF
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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 ex?racland said:
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).
FAQ: What are the solutions for e^z = 1 and e^z = 1 + i?
What is the general solution for e^z = 1?
The general solution for e^z = 1 is z = 2πni, where n is any integer. This means that any complex number z that satisfies the equation must be a multiple of 2πi.
Can you provide an example of a specific solution for e^z = 1?
One specific solution for e^z = 1 is z = 0, since e^0 = 1. Another example is z = 2πi, since e^(2πi) = cos(2π) + isin(2π) = 1.
How do you prove that z = 2πni is a solution for e^z = 1?
To prove that z = 2πni is a solution for e^z = 1, we can use Euler's formula e^(ix) = cos(x) + isin(x). Plugging in z = 2πni, we get e^(2πni) = cos(2πn) + isin(2πn) = 1. Therefore, z = 2πni satisfies the equation e^z = 1.
Are there any other solutions for e^z = 1 besides z = 2πni?
Yes, there are infinite solutions for e^z = 1. In addition to z = 2πni, any complex number that can be expressed as z = 2πni + 2kπi, where k is any integer, is also a solution. This is because e^(2πni + 2kπi) = e^(2πni)e^(2kπi) = 1 * 1 = 1.
Can you graphically represent the solutions of e^z = 1?
Yes, the solutions of e^z = 1 can be represented on the complex plane as a line passing through the origin with a slope of 2π. This line crosses through all the solutions, which are the points where the line intersects with the imaginary axis.