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wie wandle ich |z−1|+ z = 3−i (wobei i die imaginäre einheit ist) in z=(7/4)-i um

1. Formel: 

$\begin{array}{|rcll|} \hline z &=& a+bi \\ |z| &=& \sqrt{a^2+b^2} \\ \hline \end{array}$

2. |z-1|:

$\begin{array}{|rcll|} \hline |z-1| = |a+bi-1| &=& |(a-1)+bi| \\ &=& \sqrt{(a-1)^2+b^2} \\ &=& \sqrt{a^2-2a+1+b^2} \\ \hline \end{array}$

3. Berechnung von a und b:

$\begin{array}{|rcll|} \hline |z-1|+z &=& 3-i \\\\ \overbrace{\underbrace{\sqrt{a^2-2a+1+b^2} +a}_{=3}}^{Re(z)}+\overbrace{\underbrace{b}_{=-1}\cdot i}^{Im(z)} &=& 3-1\cdot i \\\\ \Rightarrow \mathbf{b} & \mathbf{=} & \mathbf{-1} \\\\ \Rightarrow \sqrt{a^2-2a+1+b^2} +a &=&3 \quad &| \quad b=-1\\ \sqrt{a^2-2a+1+(-1)^2} +a &=& 3 \\ \sqrt{a^2-2a+1+1} +a &=& 3 \\ \sqrt{a^2-2a+2} +a &=& 3 \quad &| \quad - a\\ \sqrt{a^2-2a+2} &=& 3-a \quad &| \quad \text{quadriere beide Seiten} \\ a^2-2a+2 &=& (3-a)^2 \\ \not{a^2}-2a+2 &=& 9-6a+\not{a^2} \\ -2a+2 &=& 9-6a \quad &| \quad +6 a\\ 6a-2a+2 &=& 9 \quad &| \quad -2 \\ 4a &=& 9-2 \\ 4a &=& 7 \quad &| \quad :4\\ \mathbf{a} & \mathbf{=} & \mathbf{\frac74} \\ \hline \end{array}$

4. Berechnung von z:

$\begin{array}{|rcll|} \hline z &=& a+bi \\ z &=& \frac74-1\cdot i \\ \mathbf{z} & \mathbf{=} & \mathbf{\frac74- i} \\ \hline \end{array}$

5. Probe: $z = \frac74-i$

$\begin{array}{|rcll|} \hline |\frac74- i-1|+ \frac74-i &\stackrel{?} =& 3-i \\ |\frac74-1- i |+ \frac74-\not{i} &\stackrel{?} =& 3-\not{i} \\ |\frac34- i |+ \frac74 &\stackrel{?} =& 3 \quad &| \quad |\frac34- i | = \sqrt{(\frac34)^2+(-1)^2} \\ \sqrt{(\frac34)^2+(-1)^2}+ \frac74 &\stackrel{?} =& 3 \\ \sqrt{(\frac34)^2+1}+ \frac74 &\stackrel{?} =& 3 \\ \sqrt{\frac{9}{16}+\frac{16}{16}}+ \frac74 &\stackrel{?} =& 3 \\ \sqrt{\frac{25}{16}}+ \frac74 &\stackrel{?} =& 3 \\ \frac54+ \frac74 &\stackrel{?} =& 3 \\ \frac{12}{4} &\stackrel{?} =& 3 \\ 3 &=& 3 \checkmark \\ \hline \end{array}$

wie wandle ich |z−1|+ z = 3−i (wobei i die imaginäre einheit ist) in z=(7/4)-i um

$| z-1|+z=3-i$

$2z-4=-\sqrt{-1}$                      quadrieren

$4z^2-16z+16=-1$     rechte Seite = 0

$4z^2-16z+17=0$

a           b           c

$z = {-b \pm \sqrt{b^2-4ac} \over 2a}$                a, b, c einsetzen

$\large z = {16 - \sqrt{256-4\times 4\times 17} \over 8}$

$z= \frac{16-\sqrt{-16}}{8}$

$\large z= \frac{16-4i}{8}$

$\large z=\frac{4 - i}{2}$

$\large z=2-\frac{i}{2}$

Probe

$| z-1|+z=3-i$

$| 2-\frac{i}{2}-1|+2-\frac{i}{2}=3-i$

$3-i=3-i$

  !

|z−1|+ z = 3−i

z=(7/4)-i              unrichtig!

$\frac{7}{4}-1+\frac{7}{4}-i=3-i$

$\frac{14}{4}-1-i=3-i$

$(\frac{10}{4}-i)\neq (3-i)$  q e d

  !