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\,\textrm{.}

We know already that this equation has roots

\displaystyle w=\left\{\begin{align} -i\,,&\\[5pt] i\,,& \end{align}\right.

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

\displaystyle \frac{z+i}{z-i}=-i\quad or \displaystyle \quad\frac{z+i}{z-i}=i\,\textrm{.}

We solve these equations one by one.

• \displaystyle (z+i)/(z-i)=-i:

Multiply both sides by \displaystyle z-i,

\displaystyle z+i=-i(z-i)\,\textrm{.}

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\,\textrm{.}

This gives

\displaystyle z = \frac{-1-i}{1+i} = \frac{-(1+i)}{1+i} = -1\,\textrm{.}

• \displaystyle (z+i)/(z-i)=i:

Multiply both sides by \displaystyle z-i,

\displaystyle z+i=i(z-i)\,\textrm{.}

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\,\textrm{.}

This gives

\displaystyle z = \frac{1-i}{1-i} = 1\,\textrm{.}

The solutions are therefore \displaystyle z=-1 and \displaystyle z=1\,.