# Thread: Complex number proof question

2. ## Re: Complex number proof question

Let $z = a + ib$ where $a^2+b^2 \neq 0$

\begin{aligned} \left|\frac{1}{z}\right| &= \left|\frac{a-ib}{a^2+b^2}\right| \\ &= \sqrt{\frac{a^2}{(a^2+b^2)^2} + \frac{b^2}{(a^2+b^2)^2}} \\ &= \sqrt{\frac{a^2+b^2}{(a^2+b^2)^2}} \\&= \sqrt{\frac{1}{a^2+b^2}} = \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{|z|}\end{aligned}

3. ## Re: Complex number proof question

Originally Posted by fan96
Let $z = a + ib$ where $a^2+b^2 \neq 0$

\begin{aligned} \left|\frac{1}{z}\right| &= \left|\frac{a-ib}{a^2+b^2}\right| \\ &= \sqrt{\frac{a^2}{(a^2+b^2)^2} + \frac{b^2}{(a^2+b^2)^2}} \\ &= \sqrt{\frac{a^2+b^2}{(a^2+b^2)^2}} \\&= \sqrt{\frac{1}{a^2+b^2}} = \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{|z|}\end{aligned}
Thank you.
But I don't get what you did in the second line of working.

4. ## Re: Complex number proof question

\begin{aligned}\left|\frac{a-ib}{a^2+b^2}\right| &= \left|\frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2}\right|\\ &=\sqrt{\left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2}\end{aligned}

5. ## Re: Complex number proof question

Ah, that clears it up. Thanks alot!

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