I wrote about the book here.
We discuss problems no. ##3, 4, page 21 (solved on page 177) and #38, page 25 (solved on page 185).
Problem #3 "Let $\varepsilon$ be a complex cube root of unity. Prove that
$|z-1|^2+|z-\varepsilon|^2+|z-\varepsilon^2|^2=3(|z|^2+1)"$
Problem #4 "Let $\varepsilon$ be a complex cube root of unity. Prove that
$|z+u|^2+|z+\varepsilon u|^2+|z+\varepsilon^2 u|^2=3(|z|^2+|u|^2)$
whatever $z,\;u\in\mathbb{C}$ are."
Problem #38 "Prove that :
a) $(z-1)^2+(z-\varepsilon)^2+(z-\bar\varepsilon)^2=3z^2$
where $\varepsilon=\cos\frac{2\pi}{3}+\imath \sin\frac{2\pi}{3};$
$\textbf{b)}\;\;\;(z-1)^2+(z-\varepsilon)^2+(z-\varepsilon ^2)^2+\dots+(z-\varepsilon ^{n-1})^2=nz^2$
where $\varepsilon=\cos \frac{2\pi}{n}+\imath \sin \frac{2\pi}{n};$
$$\textbf{c)}\;\;\; \sum_{k=0}^{n-1}|z-\varepsilon ^k|^2=n(|z|^2+1|)."$$
Solution of #4(CiP - Same as the solution on page 177)
The cube roots of unity are $\left \{1,\;-\frac{1}{2}\pm \imath \frac{\sqrt{3}}{2}\right \}$. Let $\varepsilon \neq 1$ be one of them. We have
$\varepsilon ^3=1,\;\;\;1+\varepsilon+\varepsilon ^2=0,\;\;\;\bar \varepsilon =\varepsilon ^2 \;\;\;\varepsilon \cdot \bar \varepsilon =1\tag{U3}$
Using that $|z|^2=z \cdot \bar z,\;\;(\forall) z\in \mathbb{C}$ we have
$|z+u|^2+|z+\varepsilon u|^2+|z+\varepsilon ^2 u|^2=$
$=(z+u)\cdot \overline{(z+u)}+(z+\varepsilon u)\cdot \overline{(z+\varepsilon u)}+(z+\varepsilon ^2 u)\cdot \overline{(z+\varepsilon ^2 u)}=$
$=(z+u)(\bar z +\bar u)+(z+\varepsilon u)(\bar z+\bar \varepsilon \bar u)+(z+\varepsilon ^2 u)(\bar z+\overline{\varepsilon ^2}\bar u)\overset{(U_3)}{=}$
$=(z\bar z+z\bar u+\bar z u+u\bar u)+(z \bar z+\varepsilon ^2 z \bar u+\varepsilon \bar z u+u \bar u)+(z\bar z+\varepsilon z \bar u+\varepsilon ^2 \bar z u+\varepsilon ^3 u \bar u)=$
$=3|z|^2+z\bar u (1+\varepsilon^2+\varepsilon)+\bar z u (1+\varepsilon +\varepsilon^2)+3|u|^2\underset{(U_3)}{=}3|z|^2+3|u|^2.$
$\blacksquare$
Solution of #3(CiP)
If $u=-1$, the statement of Problem #3 is obtained, or the calculation can be redone in this particular case.
$\blacksquare$
REMARK CiP On the cited page 177, the author of the book mentions another possible solution, using the identity(according to page 41, Exercise #10a)) :
$|z_1+z_2|^2+|z_2+z_3|^2+|z_3+z_1|^2=|z_1|^2+|z_2|^2+|z_3|^2+|z_1+z_2+z_3|^2$
Damn, I couldn't do this.
<end REM>
Solution of #38(CiP)
a) We have
$(z-1)^2+(z-\varepsilon)^2+(z-\bar \varepsilon)^2 \overset{(U_3)}{=}(z-1)^2+(z-\varepsilon)^2+(z-\varepsilon^2)^2=$
$=(z^2-2z+1)+(z^2-2\varepsilon z+\varepsilon^2)+(z^2-2\varepsilon^2z+\varepsilon^4)=$
$=3z^2-2(1+\varepsilon+\varepsilon^2)z+(1+\varepsilon^2+\varepsilon)\underset{(U_3)}{=}3z^2$
$\blacksquare$
b) Let $\varepsilon=\cos\frac{2\pi}{n}+\imath \sin \frac{2\pi}{n}$; we have $\varepsilon ^n=1$, and since $1-\varepsilon ^n=(1-\varepsilon)(1+\varepsilon+\varepsilon^2+\dots+\varepsilon ^{n-1})=0$ it follows
$$\sum_{k=0}^{n-1}\varepsilon ^k=0. \tag{1}$$
But we also have
$$\sum_{k=0}^{n-1}(\varepsilon ^k)^2=0;\tag{2}$$
this is verified by distinguishing between odd and even cases for $n$ :
$$n=2m+1\Rightarrow\sum_{k=0}^{2m}\varepsilon ^{2k}=\sum_{k=0}^m \varepsilon ^{2k}+\sum_{k=m+1}^{2m}\varepsilon^{2k}\;\;\;\;\;\overset{\varepsilon^{2m+1}=1}{\underset{\varepsilon^{2m+2}=\varepsilon^1,\;\varepsilon^{2m+4}=\varepsilon^3,\dots, \varepsilon^{4m}=\varepsilon^{2m-1}}{=}}\;\;\sum_{l=0}^{2m}\varepsilon^l\overset{(1)}{=}0;$$
$$n=2m\Rightarrow\sum_{k=0}^{2m-1}\varepsilon^{2k}=\sum_{k=0}^{m-1}\varepsilon^{2k}+\sum_{k=m}^{2m-1}\varepsilon^{2k}\;\;\;\;\;\overset{\varepsilon^{2m}=1}{\underset{\varepsilon^{2m}=1,\;\varepsilon^{2m+2}=\varepsilon^2,\dots, \varepsilon^{4m-2}=\varepsilon^{2m-2}}{=}}\;\;=2\cdot \sum_{l=0}^{m-1}\varepsilon^{2l}=0$$
the latter based on identity $1-\varepsilon^{2m}=(1-\varepsilon^2)(1+\varepsilon^2+\varepsilon^4+\dots+\varepsilon^{2m-2})=0$.
So we calculate
$$\sum_{k=0}^{n-1}(z-\varepsilon^k)^2=\sum_{k=0}^{n-1}(z^2-2z\varepsilon^k+(\varepsilon^k)^2=nz^2-2z\cdot \sum_{k=0}^{n-1}\varepsilon^k+\sum_{k=0}^{n-1}(\varepsilon ^k)^2\;\;\underset{(1)\;(2)}{=}nz^2$$
$\blacksquare$
c) Let us note that, along with (1), we also have its conjugate relation :
$$\sum_{k=0}^{n-1}(\bar \varepsilon)^k=0 \tag{3}$$
We obtain by an easy calculation :
$$\sum_{k=0}^{n-1}|z-\varepsilon^k|^2=\sum_{k=0}^{n-1}(z-\varepsilon^k)\cdot \overline{(z-\varepsilon^k)}=\sum_{k=0}^{n-1}(z-\varepsilon^k)(\bar z-\bar \varepsilon ^k)=$$
$$=\sum_{k=0}^{n-1}(z\cdot \bar z-\bar z \cdot \varepsilon^k-z\cdot \bar \varepsilon^k+\varepsilon^k\cdot \bar \varepsilon^k)=$$
$$\overset{z\cdot \bar z=|z|^2}{\underset{\varepsilon \cdot \bar \varepsilon=|\varepsilon|^2=1}{=}}\;\;\;\sum_{k=0}^{n-1}(|z|^2+1)-\bar z\cdot \sum_{k=0}^{n-1}\varepsilon^k-z\cdot \sum_{k=0}^{n-1}\bar \varepsilon^k\overset{(1)}{\underset{(3)}{=}}n(|z|^2+1).$$
REMARK CiP These formulas take place starting from $n=2$ :
$|z-1|^2+|z+1|^2=2(|z|^2+1);$
$n=3$ is the subject of exercise #3.
$\blacksquare$
