Complex number proof question (1 Viewer)

fan96 · Feb 11, 2018

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

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

Karldahemster · Feb 11, 2018

fan96 said:
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.

fan96 · Feb 11, 2018

Karldahemster · Feb 11, 2018

Ah, that clears it up. Thanks alot!

Complex number proof question (1 Viewer)

Karldahemster

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fan96

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Karldahemster

Student

fan96

617 pages

Karldahemster

Student

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