Proving a complex no

**Tutu** · November 17th 2012, 06:22 AM

How do I prove that

|z/w| = |z|/|w| ?

I had let z = a+bi and w = c+di so a+bi/c+di
= a/c + (b/d)i
So modulus is sqrt( (a/c)^2 + (b/d)^2)
And then..I have a feeling I've gone really wrong here.

Thank you very much for your help and time!

**veileen** · November 17th 2012, 06:57 AM

Uhm... O.o

If $z=a+ib$ and $w=c+id$ , then $\frac{a+ib}{c+id}$ is not necessarily equal with $\frac{a}{c}+\frac{b}{d}i$ .

$\frac{z}{w}=\frac{a+ib}{c+id}=\frac{(a+ib)(c-id)}{c^2+d^2}=\frac{ac+bd+(bc-ad)i}{c^2+d^2}$

$\Rightarrow \left | \frac{z}{w} \right |=\sqrt{\left ( \frac{ac+bd}{c^2+d^2}^{} \right )^2+\left ( \frac{bc-ad}{c^2+d^2} \right )^2}=\sqrt{\frac{a^2c^2+b^2d^2+b^2c^2+a^2d^2}{(c^2 +d^2)^{2}}}$

$=\sqrt{\frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2)^ 2}}=\sqrt{\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}}= \sqrt{\frac{a^2+b^2}{c^2+d^2}}=\frac{\sqrt{a^2+b^2 }}{\sqrt{c^2+d^2}}=\frac{|z|}{|w|}$