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Mathematics-Online course: Prepcourse Mathematics - Basics - Complex Numbers

Division of Complex Numbers

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The quotient $ z_1/z_2$ of two complex numbers

$\displaystyle z_k = x_k + \mathrm{i} y_k = r_k \exp(\mathrm{i}\varphi_k) $

is

$\displaystyle \frac{x_1x_2+y_1y_2}{x_2^2+y_2^2} + \frac{x_2y_1-x_1y_2}{x_2^2+y_2^2}\,\mathrm{i} = \frac{r_1}{r_2}\exp(\mathrm{i}(\varphi_1-\varphi_2))\, . $

In particular,

$\displaystyle \frac{1}{z} = \frac{1}{r^2} \bar z = \frac{1}{r} \exp(-\mathrm{i}\varphi) = \frac{x}{r^2} - \frac{y}{r^2}\,\mathrm{i} \,. $

A complex number's reciprocal can be constructed via reflection at the unit circle $ C$, as is illustrated in the following figure.

\includegraphics[height=6cm]{a_division_bild}

The complex conjugate $ w = \bar z$ is the intersection of the diagonals of the quadrilateral formed by the tangents at $ C$ passing through $ z$ and the perpendicular radii. The number $ z$ is then obtained by reflection at the real axis.

(Authors: Höllig/Kopf/Abele)
With $ z_k = x_k + \mathrm{i} y_k=r_k \exp (i \varphi_k)$, the quotient of two complex numbers can be calculated as follows:
$\displaystyle \frac{z_1}{z_2}$ $\displaystyle =$ $\displaystyle \frac{x_1+\mathrm{i} y_1 }{x_2 + \mathrm{i} y_2 } = \frac{(x_1 +\mathrm{i}y_1)(x_2-\mathrm{i}y_2)} {(x_2+\mathrm{i}y_2)(x_2-\mathrm{i}y_2)}$  
  $\displaystyle =$ $\displaystyle \frac{(x_1x_2+y_1y_2) + (x_2y_1 - x_1y_2)\mathrm{i}} {x_2^2+y_2^2}$  

or in polar coordinates via

$\displaystyle \frac{z_1}{z_2} = \frac{r_1 \exp (\mathrm{i} \varphi_1)}{r_2 \ex... ...2)} = \frac{r_1}{r_2} \exp (\mathrm{i} \varphi_1 - \mathrm{i} \varphi_2) \,. $

In particular, it follows that

$\displaystyle \frac{1}{z}$ $\displaystyle =$ $\displaystyle \frac{1}{x + \mathrm{i} y} = \frac {x - \mathrm{i} y}{(x + \mathrm{i} y)(x - \mathrm{i}y)}$  
  $\displaystyle =$ $\displaystyle \frac{\bar {z}}{x^2+y^2} = \frac{\bar{z}}{r^2} = \frac{1}{r} \ \exp(-\mathrm{i} \varphi) \,.$  

The geometric construction is based on the theorem of Pythagoras which implies

$\displaystyle \vert w\vert\,\vert z\vert = 1^2 \,, $

i.e., that $ w$ has the correct modulus. Reflection at the real line changes the sign of the argument, so that

$\displaystyle \operatorname{arg} \bar w = \operatorname{arg} (1/z) $

as claimed.

(Authors: Höllig/Kopf/Abele)
In order to calculate

$\displaystyle \frac{(1+\sqrt{3}\mathrm{i})+2 \exp (-\mathrm{i}\pi/6)}{\exp(\mathrm{i}\pi/2) (1-\mathrm{i})}\, , $

we form the sum in the numerator in standard representation,

$\displaystyle (1+\sqrt{3}\mathrm{i})+ 2\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\mathrm{i} \right) = (1+\sqrt{3})+(\sqrt{3}-1)\mathrm{i}\, , $

and multiply the factors of the denominator in polar coordinates,

$\displaystyle \exp(\mathrm{i}\pi/2)\cdot \sqrt{2}\exp(-\mathrm{i}\pi/4) = \sqrt{2}\exp(\mathrm{i}\pi/4) = 1 + \mathrm{i}\, . $

Thus the quotient is

$\displaystyle \frac{((1+\sqrt{3})+(\sqrt{3}-1)\mathrm{i}) (1-\mathrm{i})}{(1+\m... ...mathrm{i})} = \frac{2\sqrt{3} - 2\mathrm{i}}{2} = 2 \exp(-\mathrm{i}\pi/6)\, , $

which transforms to the standard representation as

$\displaystyle 2(\cos(\pi/6)-\mathrm{i}\sin(\pi/6)) = \sqrt{3}-\mathrm{i} \,. $

(Authors: Höllig/Kopf/Abele)
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  automatically generated 10/23/2009