1. ## Complex Analysis Problem

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 z1 be the argument of z+1, and x2 be the argument of z-1
|z+1|100e100x1i=|z-1|100e100x2i
Taking the 100-th root:
|z+1|e(100x1+2πk)i/100=|z-1|e(100x2+2πk)i/100
|z+1|ex1i+50πki=|z-1|ex2i+50πki
|z+1|ex1i=|z-1|ex2i
|z+1|[cos(x1)+isin(x1)]=|z-1|[cos(x2)+isin(x2)]
|z+1|cos(x1)-|z-1|cos(x2)=i[|z-1|sin(x2)-|z+1|sin(x1)]

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

2. ## Re: Complex Analysis Problem

Hi flybynight!

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 z1 be the argument of z+1, and x2 be the argument of z-1
|z+1|100e100x1i=|z-1|100e100x2i
Taking the 100-th root:
|z+1|e(100x1+2πk)i/100=|z-1|e(100x2+2πk)i/100

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.

|z+1|ex1i+50πki=|z-1|ex2i+50πki
|z+1|ex1i=|z-1|ex2i

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

|z+1|[cos(x1)+isin(x1)]=|z-1|[cos(x2)+isin(x2)]
|z+1|cos(x1)-|z-1|cos(x2)=i[|z-1|sin(x2)-|z+1|sin(x1)]

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:

$|z + 1| = |z - 1|$

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. $\blacksquare$

3. ## 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?

4. ## Re: Complex Analysis Problem

Originally Posted by flybynight
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?
You have:

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

For two such numbers to be the same, the lengths have to be the same and the angles have to be the same (modulo 2pi).

Take the length left and right:

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

Each imaginary number can be represented in the form $r e^{i\phi}$, where r is the length and phi is the argument, which are both real numbers.
Raising it to the power 100 gives: $r^{100} e^{100i\phi}$.
The length is $r^{100}$.
Since they are supposed to be the same, the lengths without the power will also have to be the same.

In other words:

$|z+1| = |z-1|$