Solve e^z=1: Complex Solutions

[wany]

Homework Statement

solve e^z=1

Homework Equations

e^z=e^xe^{iy}

The Attempt at a Solution

So let 1=r cis \theta
so then
e^x=r
e^{iy}=cis \theta
Then this happens when y=2k\pi i for k=...,-1,0,1,...
Then x=log(r).

I know I am missing something.

 

[Char. Limit]

Homework Statement

solve e^z=1

Homework Equations

e^z=e^xe^{iy}

The Attempt at a Solution

So let 1=r cis \theta
so then
e^x=r
e^{iy}=cis \theta
Then this happens when y=2k\pi i for k=...,-1,0,1,...
Then x=log(r).

I know I am missing something.

Doesn't r have to equal 1 here?

 

[Dick]

r=1. Is that what you are missing?

 

[wany]

Why does r have to equal 1? Or is this because r is the radius and therefore it must be 1?
Then since r=1, x=0 since ln(1)=0.

 

[Char. Limit]

I thought of a better way to do this. 1 = e^(2 pi i n) for all integer n, right?

So, e^(z) = e^(2 pi i n), and it's reasonable to conclude that z=2*pi*i*n.

 

[Dick]

Why does r have to equal 1? Or is this because r is the radius and therefore it must be 1?
Then since r=1, x=0 since ln(1)=0.

It's because |cis(y)|=1. e^x*cis(y)=1 so |e^x|*|cis(y)|=1 means |e^x| must also be 1.

 

[wany]

Alright thank you very much for your help. That makes sense. Have a great weekend.

 

[HallsofIvy]

Or, another way of looking at it: in the complex plane, 1 is on the "real" axis, at a distance 1 from 0, at an angle, with the x-axis, of 0 (or, more generally, 2n\pi. That's why 1= 1e^{2n\pi}= 1(cos(2n\pi)+ i sin(2n\pi))= 1 cis(2n\pi).