Aus Online Mathematik Brückenkurs 2

Let us first divide both sides by \displaystyle 4+i, so that the coefficient in front of \displaystyle z^2 becomes \displaystyle 1,

\displaystyle z^2 + \frac{1-21i}{4+i}z = \frac{17}{4+i}\,\textrm{.}

The two complex quotients become

\displaystyle \begin{align} \frac{1-21i}{4+i} &= \frac{(1-21i)(4-i)}{(4+i)(4-i)} = \frac{4-i-84i+21i^2}{4^2-i^2}\\[5pt] &= \frac{-17-85i}{16+1} = \frac{-17-85i}{17} = -1-5i\,,\\[10pt] \frac{17}{4+i} &= \frac{17(4-i)}{(4+i)(4-i)} = \frac{17(4-i)}{4^2-i^2}\\[5pt] &= \frac{17(4-i)}{17} = 4-i\,\textrm{.} \end{align}

Thus, the equation becomes

\displaystyle z^2 - (1+5i)z = 4-i\,\textrm{.}

Now, we complete the square of the left-hand side,

\displaystyle \begin{align} \Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1+5i}{2}\Bigr)^2 &= 4-i\,,\\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \Bigl(\frac{1}{4}+\frac{5}{2}\,i+\frac{25}{4}i^2 \Bigr) &= 4-i\,,\\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 - \frac{1}{4} - \frac{5}{2}i + \frac{25}{4} &= 4-i\,, \\[5pt] \Bigl(z-\frac{1+5i}{2}\Bigr)^2 &= -2+\frac{3}{2}\,i\,\textrm{.} \end{align}

If we set \displaystyle w=z-\frac{1+5i}{2}, we have a binomial equation in \displaystyle w,

\displaystyle w^2 = -2+\frac{3}{2}\,i

which we solve by putting \displaystyle w=x+iy,

\displaystyle (x+iy)^2 = -2+\frac{3}{2}\,i

or, if the left-hand side is expanded,

\displaystyle x^2-y^2+2xyi = -2+\frac{3}{2}\,i\,\textrm{.}

If we identify the real and imaginary parts on both sides, we get

\displaystyle \begin{align} x^2-y^2 &= -2\,,\\[5pt] 2xy &= \frac{3}{2}\,, \end{align}

and if we take the magnitude of both sides and put them equal to each other, we obtain a third relation:

\displaystyle x^2 + y^2 = \sqrt{(-2)^2+\bigl(\tfrac{3}{2}\bigr)^2} = \tfrac{5}{2}\,\textrm{.}

Strictly speaking, this last relation is contained within the first two and we include it because makes the calculations easier.

Together, the three relations constitute the following system of equations:

\displaystyle \left\{\begin{align} x^2-y^2 &= -2\,,\\[5pt] 2xy &= \frac{3}{2}\,,\\[5pt] x^2 + y^2 &= \frac{5}{2}\,\textrm{.} \end{align} \right.

From the first and the third equations, we can relatively easily obtain the values that \displaystyle x and \displaystyle y can take.

Add the first and third equations,

	\displaystyle x^2	\displaystyle {}-{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle -2
\displaystyle +\ \	\displaystyle x^2	\displaystyle {}+{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle \tfrac{5}{2}
	\displaystyle 2x^2			\displaystyle {}={}	\displaystyle \tfrac{1}{2}

which gives that \displaystyle x=\pm \tfrac{1}{2}.

Then, subtract the first equation from the third equation,

	\displaystyle x^2	\displaystyle {}+{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle \tfrac{5}{2}
\displaystyle -\ \	\displaystyle \bigl(x^2	\displaystyle {}-{}	\displaystyle y^2	\displaystyle {}={}	\displaystyle -2\rlap{\bigr)}
			\displaystyle 2y^2	\displaystyle {}={}	\displaystyle \tfrac{9}{2}

i.e. \displaystyle y=\pm\tfrac{3}{2}.

This gives four possible combinations,

\displaystyle \left\{\begin{align} x &= \tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= \tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align} \right.

of which only two also satisfy the second equation.

\displaystyle \left\{\begin{align} x &= \tfrac{1}{2}\\[5pt] y &= \tfrac{3}{2} \end{align}\right. \qquad\text{and}\qquad \left\{\begin{align} x &= -\tfrac{1}{2}\\[5pt] y &= -\tfrac{3}{2} \end{align}\right.

This means that the binomial equation has the two solutions,

\displaystyle w=\frac{1}{2}+\frac{3}{2}\,i\qquad and \displaystyle \qquad w=-\frac{1}{2}-\frac{3}{2}\,i\,,

and that the original equation has the solutions

\displaystyle z=1+4i\qquad and \displaystyle \qquad z=i

according to the relation \displaystyle w=z-\frac{1+5i}{2}.

Finally, we check that the solutions really do satisfy the equation.

\displaystyle \begin{align} z=1+4i:\quad (4+i)z^2+(1-21i)z &= (4+i)(1+4i)^2+(1-21i)(1+4i)\\[5pt] &= (4+i)(1+8i+16i^2)+(1+4i-21i-84i^2)\\[5pt] &= (4+i)(-15+8i)+1-17i+84\\[5pt] &= -60+32i-15i+8i^2+1-17i+84\\[5pt] &= -60+32i-15i-8+1-17i+84\\[5pt] &= 17\,,\\[10pt] z={}\rlap{i:}\phantom{1+4i:}{}\quad (4+i)z^2+(1-21i)z &= (4+i)i^2 + (1-21i)i\\[5pt] &= (4+i)(-1)+i-21i^2\\[5pt] &= -4-i+i+21\\[5pt] &= 17\,\textrm{.} \end{align}