If x+iy=(u+iv)^n , where n is an integer and x,y,u,v∈mathbbR…

If $x+iy=(u+iv)^n$, where $n$ is an integer and $x,y,u,v\in\mathbb{R}$, prove that $x^2+y^2=(u^2+v^2)^n$.

Step 1 · Reduce to a modulus statement

Let $z=u+iv$, so $x+iy=z^n$. Taking moduli of both sides and squaring will give the result, since $|x+iy|^2=x^2+y^2$ and $|z|^2=u^2+v^2$.

Step 2 · Prove $|z^n|=|z|^n$ for positive integers by induction

Base case ($n=1$): $|z^1|=|z|^1$, trivially true.

Inductive step: suppose $|z^k|=|z|^k$. Using $|z_1z_2|=|z_1||z_2|$ (Problem 1, part $1^{\circ}$):

$$ |z^{k+1}|=|z^k\cdot z|=|z^k||z|=|z|^k|z|=|z|^{k+1} $$

so $|z^n|=|z|^n$ holds for every positive integer $n$.

Step 3 · Extend to negative integers and $n=0$

For $n=0$ (and $z\neq0$): $z^0=1$, so $|z^0|=1=|z|^0$.

For $n=-m$ with $m>0$:

$$ |z^{-m}|=\left|\frac1{z^m}\right|=\frac1{|z^m|}=\frac1{|z|^m}=|z|^{-m} $$

So $|z^n|=|z|^n$ holds for every integer $n$ (with $z\neq0$ when $n\leq0$).

Step 4 · Apply to the given equation

From $x+iy=(u+iv)^n$, taking moduli:

$$ |x+iy|=|(u+iv)^n|=|u+iv|^n $$

Squaring both sides:

$$ \boxed{x^2+y^2=(u^2+v^2)^n} $$

Step 5 · Connection to polar form

Writing $u+iv=re^{i\theta}$ with $r=\sqrt{u^2+v^2}$, De Moivre's formula gives $(u+iv)^n=r^n e^{in\theta}$, whose modulus is $r^n=(u^2+v^2)^{n/2}$; squaring recovers $x^2+y^2=(u^2+v^2)^n$ directly.

Step 6 · Numerical check

Let $u=1,v=1,n=2$: $(1+i)^2=2i$, so $x=0,y=2$, and $x^2+y^2=4$. Also $(u^2+v^2)^n=(1+1)^2=4$. ✓