Aus Online Mathematik Brückenkurs 2

If we treat the expression \displaystyle w=\frac{z+i}{z-i} as an unknown, we have the equation

\displaystyle w^{2}=-1

We know already that this equation has roots

\displaystyle w=\left\{ \begin{array}{*{35}l} -i \\ i \\ \end{array} \right.

so \displaystyle z\text{ } should satisfy one of the equation's

\displaystyle \frac{z+i}{z-i}=-i or \displaystyle \frac{z+i}{z-i}=i

We solve these equations one by one.

\displaystyle \underline{\underline{\frac{z+i}{z-i}=-i}}

Multiply both sides by \displaystyle z-i:

\displaystyle z+i=-i\left( z-i \right)

Move all the \displaystyle z -terms over to the left-hand side and all the constants to the right-hand side,

\displaystyle z+iz=-1-i

This gives

\displaystyle z=\frac{-1-i}{1+i}=\frac{-\left( 1+i \right)}{1+i}=-1

\displaystyle \underline{\underline{\frac{z+i}{z-i}=i}}

Multiply both sides by \displaystyle z-i:

\displaystyle z+i=i\left( z-i \right)

Move all the z-terms over to the left-hand side and all the constants to the right-hand side,

\displaystyle z-iz=1-i

This gives

\displaystyle z=\frac{1-i}{1-i}=1

The solutions are therefore \displaystyle z=-\text{1} and \displaystyle z=\text{1}.