How many MISTAKES are allowed in a book ? // Hoeveel fouten zijn toegestaan ​​in een boek ?

 Answer: NOT TOO MUCH

Let's take the first problem from a book.

Page 21, Chapter 1 EQUALITIES, Section 1.1 Equalities of one complex variable, Paragraph 1.A, Problem  #1 (in translation)

               "Let  $z\in\mathbb{C}$ such that  $z^2=-3+4\imath$. Calculate:

                       a)  $A=z^2+\bar z^2;$

                       b)  $B=\left | z-\frac{1}{\bar z } \right |;$

                       c)  $C=z+\frac{1}{\bar z};$

                       d)  $D=Re\;z+Im\;z.$"

Do you see the mistakes ?

ANSWER CiP

A. $-6$  ; B. $\frac{4\sqrt{5}}{5}$  ; C. $\pm\frac{6}{5}(1+2\imath)$  ;  D. $\pm 3$.

In reality  $z=\pm(1+2\imath)$

                         Solution CiP

               Let  $z=x+\imath y$. From  $-3+4\imath=z^2=(x+\imath y)^2=(x^2-y^2)+2xy \cdot \imath$

it follow  $x^2-y^2=-3,\;\;2xy=4$  so  $y=\frac{2}{x}\;\;and\;\;x^2-\frac{4}{x^2}=-3$.  Hence

$x^2=1$  and therefore  $z_{1,2}=\pm(1+2\imath).$

          Regardless of the possible values ​​for  $z$, we have the calculations :

$A=z^2+\overline {z^2}=(-3+4\imath)+\overline {(-3+4\imath)}=(-3+4\imath)+(-3-4\imath)=-6;$

     We then have  $|z^2|=|-3+4\imath|=\sqrt{(-3)^2+4^2}=5=|z|^2=|\bar z|^2$  so on one side  $|\bar z|=\sqrt{5}$, and

$B=\left | \frac{z\cdot \bar z-1}{\bar z} \right |=\left | \frac{|z|^2-1}{\bar z} \right |=\left | \frac{|z^2|-1}{\bar z} \right |=\left | \frac{5-1}{\bar z}\right |=\frac{4}{|\bar z|}=\frac{4}{\sqrt{5}};$

$C=\frac{z \cdot \bar z+1}{\bar z}=\frac{|z|^2+1}{\bar z}=\frac{5+1}{\bar z}=\frac{6\cdot z}{          \bar z \cdot z}=\frac{6z}{|z|^2}=\frac{6}{5}\cdot z\;\tag{1}$

     Finally, we still need

$\bar z=\frac{5}{z}$  and  $\frac{1}{z}=\frac{z}{-3+4\imath}$

to calculate  $D=Re\;z+Im\;z=\frac{z+\bar z}{2}+\frac{z-\bar z}{2\imath}=\frac{z+\frac{5}{z}}{2}+\frac{z-\frac{5}{z}}{2\imath}=\frac{z^2+5}{2z}+\frac{z^2-5}{2\imath z}=$

$\overset{z^2=-3+4\imath}{=}\;\;\frac{2+4\imath}{2z}+\frac{-8+4\imath}{2\imath z}=\frac{1}{z}+\frac{2\imath}{z}+\frac{8\imath ^2}{2\imath z}+\frac{2}{z}=\frac{3+6\imath}{z}=$

$=3(1+2\imath) \cdot \frac{1}{z}=3(1+2\imath) \cdot \frac{z}{-3+4\imath}=3(1+2\imath)\cdot \frac{(-3-4\imath)z}{25}=\frac{3(5-10\imath)}{25}\cdot z=\frac{3}{5}(1-2\imath)\cdot z \tag{2}$

Substituting the values  $z_{1,2}$  ​​into (1) and (2) we obtain the answer.

$\blacksquare$

          Remark CiP  Compare with the official solution :