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:

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

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
[tex]\frac{1}{a+ bi}[/itex] <br /> and "rationlize the denominator". Multiply both numerator and denominator by a- bi:<br /> [tex]\frac{1}{a+ bi}\frac{a- bi}{a- bi}= \frac{a- bi}{a^2+ b^2}[/itex]<br /> which is, of course, exactly micromass's<br /> $$\frac{\overline{z}}{|z|^2}$$.<br /> <br /> That is the formula, which you say you already know. The "proof" (that that formula is correct) is just to multiply:<br /> $$(a+ bi)\frac{a- bi}{a^2+ b^2}= \frac{(a+ bi)(a- bi)}{a^2+ b^2}= \frac{a^2+ b^2}{a^2+ b^2}= 1$$[/tex][/tex]