Multiplicative inverse of complex numbers

[sarah786]

I can't find a proof for the multiplicative inverse of complex numbers... can anybody please tell me the proof (i already know what the formula is)

 

[micromass]

If you already know the formule than you're already on the good way. So, I guess the formula you have is z^{-1}=\frac{\overline{z}}{|z|^2}.

So the only thing you need to show now is that zz^{-1}=z^{-1}z=1. Just complete the following multiplication:

zz^{-1}=\frac{z\overline{z}}{|z|^2}=...

 

[LCKurtz]

If you already know the formule than you're already on the good way. So, I guess the formula you have is z^{-1}=\frac{\overline{z}}{|z|^2}.

So the only thing you need to show now is that zz^{-1}=z^{-1}z=1. Just complete the following multiplication:

zz^{-1}=\frac{z\overline{z}}{|z|^2}=...

I'll add to what Micromass has said. If you want the inverse of z = a+bi you are looking for a complex number w = x+yi such that

(a+bi)(x+yi) = 1 = 1+0i

Multiplying out the left side:

(ax - by) + (bx + ay)i = 1 + 0i

Equating real and imaginary parts:

ax - by = 1
bx + ay= 0

Solving these for x and y by determinants gives:

x = \frac{\left|\begin{array}{cc} 1 &amp; -b\\0 &amp; a \end{array}\right|}<br /> {\left|\begin{array}{cc} a &amp; -b\\b &amp; a \end{array}\right|} = \frac{a}{a^2+b^2},\, <br /> y = \frac{\left|\begin{array}{cc} a &amp; 1\\b &amp; 0 \end{array}\right|}<br /> {\left|\begin{array}{cc} a &amp; -b\\b &amp; a \end{array}\right|} = \frac{-b}{a^2+b^2}<br />

This tells you that the inverse of z is

w = \frac{a}{a^2+b^2} + \frac{-b}{a^2+b^2}i = \frac{1}{a^2+b^2}(a-bi)=\frac{\overline z}{|z|^2}

 

[HallsofIvy]

Or, just to put in my oar, to find the multiplicative inverse of a+ bi, write
\frac{1}{a+ bi}[/itex] <br /> and &quot;rationlize the denominator&quot;. Multiply both numerator and denominator by a- bi:<br /> \frac{1}{a+ bi}\frac{a- bi}{a- bi}= \frac{a- bi}{a^2+ b^2}[/itex]&lt;br /&gt; which is, of course, exactly micromass&amp;#039;s&lt;br /&gt; \frac{\overline{z}}{|z|^2}.&lt;br /&gt; &lt;br /&gt; That is the formula, which you say you already know. The &amp;quot;proof&amp;quot; (that that formula is correct) is just to multiply:&lt;br /&gt; (a+ bi)\frac{a- bi}{a^2+ b^2}= \frac{(a+ bi)(a- bi)}{a^2+ b^2}= \frac{a^2+ b^2}{a^2+ b^2}= 1