How to prove that two equations in complex numbers have no solution in common? So my complex numbers are:
$$r_1^n = 1+i$$
$$r_2^m = 2-i$$
$$r_1, r_2 \in \mathbb{C} \quad;\quad n,m \in \mathbb{N}$$
I wrote down the general $z^n=r^n(\cos(n\theta)+i\sin(n\theta))$ formula for these, but I don't know what to do after this step to prove these two equations don't have any common solution.
 A: Assume that there exists some $z$ for which $z^n=1+i$ and $z^m=2-i$ for some integers $m$ and $n$. You're right to think about converting between rectangular and polar, but in this case it's polar form that's useful:
$$1+i=\sqrt{2}e^{i\theta_1},\ \ 2-i=\sqrt{5}e^{i\theta_2}.$$
(I haven't bothered to find $\theta_1$ or $\theta_2$, since they happen not to be relevant). So, by comparing magnitudes,
$$|z|=2^{\frac{1}{2n}}=5^{\frac{1}{2m}}.$$
Can you show that this can never happen?
A: $r_1=(1+i)^{1/n}$ and and $r_2=(2-i)^{1/m}$  implies $|r_1|=\sqrt{2}^{1/n}$ and $|r_2|=\sqrt{5}^{1/m}.$. Therefore $|r_1|\neq |r_2|$ for any values of $m,n$ which intern implies that  $r_1\neq r_2$ for any values of $m,n.$
A: Suppose $n, m \in \mathbb{N}$ are such that
$$
z^n = 1 + i \\
z^m = 2 - i
$$
for some complex number $z = r e^{i\theta}$. Note that $n > 0$ and $m > 0$. Taking the absolute value of both equations we get
$$
r^n = \sqrt{2} \\
r^m = \sqrt{5}
$$
and so
$$
r^{nm} = \sqrt{2^m} \\
r^{nm} = \sqrt{5^n}.
$$
But this implies that
$$
2^m = 5^n
$$
which is impossible since the left hand side is even and the right hand side is odd.
