# If $a$ is an $n$-th root of $z$ , $b$ is an $n$-th root of $w$, and $c$ is an $n$-th root of $zw$, then $ab=c$.

How do I prove that for all natural numbers $$n$$ and complex numbers $$a, b, c, z, w$$ if $$a$$ is an $$n$$-th root of $$z$$ , $$b$$ is an $$n$$-th root of $$w$$, and $$c$$ is an $$n$$-th root of $$zw$$, then $$ab=c$$.

Thanks.

• What is the question? – blub Apr 1 at 20:33
• Perhaps you mean to say "Prove or disprove that if $a$ is an $n^{\textrm{th}}$ root of $z$ and $b$ is an $n^{\textrm{th}}$ root of $w$, then $ab$ is an $n^{\textrm{th}}$ root of $zw$"? – MPW Apr 1 at 20:35
• @blub I just added it sorry. – Sania Apr 1 at 20:35
• @Sania "Prove or disprove" is not a question, it's a command. – ajotatxe Apr 1 at 20:37

You have $$a^n=z$$ , $$b^n=w$$ and $$c^n=zw$$ so $$\left(\frac{ab}c\right)^n=1\quad$$ (the case $$z=0$$ or $$w=0$$ being trivial).
Thus all we can deduce is $$\frac{ab}c$$ is a n-root of unity, but not necessarily $$1$$.
A trivial counter example would be $$n=2, z=w=a=b=1$$ and $$c=-1$$.