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Aus Online Mathematik Brückenkurs 2

Let us first divide both sides by 4+i, so that the coefficient in front of z2 becomes 1,

z2+4+i1−21iz=174+i.

The two complex quotients become

4+i1−21i174+i=(4+i)(4−i)(1−21i)(4−i)=42−i24−i−84i+21i2=16+1−17−85i=17−17−85i=−1−5i=17(4−i)(4+i)(4−i)=42−i217(4−i)=1717(4−i)=4−i.

Thus, the equation becomes

z2−(1+5i)z=4−i.

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

z−21+5i2−21+5i2z−21+5i2−41+25i+425i2z−21+5i2−41−25i+425z−21+5i2=4−i=4−i=4−i=−2+23i.

If we set w=z−21+5i, we have a binomial equation in w,

w2=−2+23i

which we solve by putting w=x+iy,

(x+iy)2=−2+23i

or, if the left-hand side is expanded,

x2−y2+2xyi=−2+23i.

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

x2−y22xy=−2=23

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

x2+y2=(−2)2+232=25.

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:

x2−y22xyx2+y2=−2=23=25.

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}