Question

show: |z1/z2| = |z1|/|z2|

Answer 1

\[\left| Z _{1} / Z_{2} \right| = \left| Z_{1} \right| /\left| Z_{2} \right|\]

Answer 2

\(z_1,z_2\) are complex numbers?

Answer 3

ya

Answer 4

\[Z_2 \neq 0\]

Answer 5

Let \(z_1=a+bi\) and \(z_2=c+di\).

RHS:\[\frac{|z_1|}{|z_2|}=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}\]
LHS:\[\begin{align*}\left|\frac{z_1}{z_2}\right|&=\left|\frac{a+bi}{c+di}\right|\\\\
&=\left|\frac{a+bi}{c+di}\cdot\frac{c-di}{c-di}\right|\\\\
&=\left|\frac{(a+bi)(c-di)}{c^2+d^2}\right|\\\\
&=\left|\frac{ac+bd+(cb-ad)i}{c^2+d^2+0i}\right|\\\\
&=\left|\frac{\sqrt{(ac+bd)^2+(bc-ad)^2}}{\sqrt{(c^2+d^2)^2+0^2}}\right|\\\\
&=\left|\frac{\sqrt{\left(a^2c^2+2abcd+b^2d^2\right)+\left(b^2c^2-2abcd+a^2d^2\right)}}{\sqrt{(c^2+d^2)^2}}\right|\\\\
&=\left|\frac{\sqrt{a^2c^2+b^2d^2+b^2c^2+a^2d^2}}{\sqrt{(c^2+d^2)^2}}\right|\\\\
&=\left|\frac{\sqrt{a^2(c^2+d^2)+b^2(d^2+c^2)}}{\sqrt{(c^2+d^2)^2}}\right|\\\\
&=\left|\frac{\sqrt{(a^2+b^2)(c^2+d^2)}}{\sqrt{(c^2+d^2)^2}}\right|\\\\
&=\left|\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}\right|\\\\
&=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}
\end{align*}\]

Answer 6

wow thanks! I started like that but I didnt think I was going to get anywhere...

Answer 7

at least I made an error in my math i mean

Answer 8

You're welcome!