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