How to prove that two equations in complex numbers have no solution in common?

Question

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.

Answer 1

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.

Answer 2

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?

Answer 3

$r_1=(1+i)^{1/n}$ 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 $m,n.$