- #1

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I've been asked to determine the values of z that obey the equation

e^z = 1 + sqrt(3)i

I'm still not sure the concept of this question. Could someone point me in the right direction?

Thanks

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- Thread starter galipop
- Start date

- #1

- 51

- 0

I've been asked to determine the values of z that obey the equation

e^z = 1 + sqrt(3)i

I'm still not sure the concept of this question. Could someone point me in the right direction?

Thanks

- #2

matt grime

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- #3

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e^(a+bi)=(e^a)*( cos(b)+sin(b)i)

Now you need to find a and b according to the result you want.

please notice that there are infinity number of solution to that !

Best

Moshek

- #4

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So based on the first 2 replies:

e^(x+iy) = 1 + sqrt(3)i

e^x.e^iy = e^x*(cos(y)+sin(y)i) = 1 + sqrt(3)i

therefore e^x = sqrt(1*1 + sqrt(3)*sqrt(3))

so e^x = 2. which is unique.

Now I just need to find y right?

so am I on the right track?

Thanks

- #5

matt grime

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Yes, you're on the right track, you need to find the arguments.

- #6

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tan(y) = b/a

tan(y) = sqrt(3)/1

y= PI/3

you talked about infinite solutions, so then

y = PI/3 + (PI*n) where n is an integer.

so now is have

x = ln 2

and

y= PI/3 + (PI*n)

so now do I just subsitute x and y into this eqn?

e^x*(cos(y)+sin(y)i) = 1 + sqrt(3)i

- #7

HallsofIvy

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is z= log(1+ sqrt(3)) since log z= log|z|+ arg(z)+ 2πn

That is, 1+ sqrt(3)i, in "polar form" is 2 e^(2πi/3) so

z= log(2)+ 2π/3+ 2πn where n is any integer.

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