Re: Complex Analysis Problem

Hi flybynight! :)

Quote:

Originally Posted by

**flybynight** Hi everyone,

I am given this problem: show that all solutions of (z+1)^{100}=(z-1)^{100} satisfy Re(z)=0. This is what I've done to try to solve the problem:

(z+1)^{100}=(z-1)^{100}

Let z_{1} be the argument of z+1, and x_{2} be the argument of z-1

|z+1|^{100}e^{100x1i}=|z-1|^{100}e^{100x2i}

Taking the 100-th root:

|z+1|e^{(}^{100x1+2πk)i/100}=|z-1|e^{(}^{100x2+2πk)i/100}

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Here's a small mistake: the k in the right hand side does not have to be the seem as the k in the left hand side.
You should use a different letter, for instance m. } Quote:

|z+1|e^{x1i+50πki}=|z-1|e^{x2i+50πki }|z+1|e^{x1i}=|z-1|e^{x2i }

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As a consequence you can't scratch of those 50πki occurrences. } Quote:

|z+1|[cos(x_{1})+isin(x_{1})]=|z-1|[cos(x_{2})+isin(x_{2})]

|z+1|cos(x_{1})-|z-1|cos(x_{2})=i[|z-1|sin(x_{2})-|z+1|sin(x_{1})]

And if the real part is equal to the imaginary part, it seems as if both would be equal to zero, which isn't the case.

Any advice on what I did wrong or an alternate pathway to the solution?

Thanks,

Peter

For an alternate path, consider the length of the left hand side and the right hand side.

They have to be the same.

Since they are both real numbers you can take the 100-root to find:

These expressions represent the distance z has to -1, respectively the distance z has to +1.

The only values for z that have that, are the ones on the imaginary axis.

In other words: the values of z that have a real part that is zero.

Re: Complex Analysis Problem

Thanks for your help! But I don't see why the two moduli have to be the same; can you tell me a bit more on why?

Re: Complex Analysis Problem