• Let $T$ be a triangle with vertices $a, b, c$ in $\mathbb{C}$ (in anticlockwise direction)

• Then the following are Equivalent

(i) $T$ is equilateral triangle

(ii) $a+w b+w^2 c=0$, $w=c~ is ~ \frac{2 \pi}{3}$

(iii) $a^2+b^2+c^2=a b+b c+c a$

• Let $T$ be a triangle with vertices $z_1, z_4, z_3$ in $\mathbb{C}$

• If $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|$ and $z_1+z_2+z_3=0$. then show that $T$ is equilateral triangle

Given $z_1+z_2+z_3=0$

$\Rightarrow \quad z_1+z_2+z_3=0$

$\Rightarrow \quad \bar{z}_1+\bar{z}_2+\bar{z}_3=0$

divide by $\left|z_1\right|^2$, we get

$\frac{\bar{z}_1}{z_1 \bar{z}_1}+ \frac{\bar{z}_2}{z_2 \bar{z}_2}+ \frac{\bar{z}_3}{z_3 \bar{z}_3} = 0$

$\Rightarrow \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=0 \rightarrow \times z_1 z_2 z_3 \\ \Rightarrow z_2 z_3+z_1 z_3+z_1 z_2=0 \rightarrow \text { (1) }$

we have $\left(z_1+z_2+z_3\right)^2=0$

$\Rightarrow z_1^2+z_2^2+z_3^4+2\left(z_1 z_2+z_2 z_3+z_3 z_1\right)=0 \\ \Rightarrow z_1^2+z_2^2+z_3^2=0 \rightarrow \text { (2) }$

From (1)& (2)

$z_1^2+z_2^2+z_3^2 = z_1 z_2+z_2 z_3+z_3 z_1$

$\Rightarrow$ T is equilateral triangle.

• Let $z_1, z_2, z_3$ be complex numbers such that
• $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1$
• $z_1+z_2+z_3 \neq 0$
• $z_1^2+z_2^2+z_3^2=0$

Then show that for all integer $n \geq 2$

$\left|z_1^n+z_2^n+z_3^n\right| \in\{0,1,2,3\}$

observation, $z_1^2, z_2^2, z_3^2$ are distinct

$z_1^2=z_2^2 \Rightarrow \quad z_3^2=-2 z_1^2 \\ \Rightarrow\left|z_3\right|^2=2 \quad \Rightarrow \Leftarrow$

$T\left(z_1^2, z_2^2, z_3^2\right)$ is an equilateral triangle

i.e. $z_2^2=\omega z_1^2 = \omega^4 z_1^2$

$\Rightarrow z_2= \pm \omega^2 z_1$

$z_3^2=\omega^2 z_1^2$

$\Rightarrow z_3= \pm w z_1$

$\left|z_1^n+z_2^n+z_3^n\right| =\left|z_1^n\left(1 \pm \omega^{2 n} \pm \omega^n\right)\right|$

$\quad \quad =\left|1 \pm \omega^{2 n} \pm \omega^n\right| \leqslant 3$

$\omega^{2 n}=\omega^n \Leftrightarrow \omega^n=1$

$1 \quad$ $\omega$ $\quad$ $\omega^2$

$1 \quad$ $\omega$ $\quad$ $-\omega^2$

$1 \quad$ $-\omega$ $\quad$ $\omega^2$

$1 \quad$ $-\omega$ $\quad$ $-\omega^2$

Sum $\in \text{\{0,1,2,3\}}.$

Compute

$\left(1-w+w^2\right)\left(1-w^2+w^4\right) \cdots\left(1-w^n+w^{2 n}\right)$

Recall: $1+w+w^2 =0$

$\omega^n= \begin{cases}1 \quad \text { if } n \equiv 0(\operatorname{mod} 3) \\ \omega \quad \text { if } n \equiv 1(\operatorname{mod} 3) \\ \omega^2 \quad \text { if } n \geq 2(\operatorname{mod} 3)\end{cases}$

$1-w+w^2 =-w-w \ =-2 w$

$1-\omega^2+\omega^4=1-\omega^2+\omega=-2 \omega^2$

$1-\omega^3+w^6=1 \\ 1-w^n+w^{2n}= \quad \begin{cases} 1 \text { if } n \equiv 0(\bmod 3) \\ -2 \omega \text { if } n \equiv 1(\bmod 3) \\ -2 \omega^2 \text { if } n \equiv 2(\bmod 3) \end{cases}$

The 1st three term product $=(-2 \omega)\left(-2 \omega^2\right)\cdot 1=2^2$

$\left(1-\omega+\omega^2\right)\left(1-\omega^2+\omega^4\right) \cdots\left(1-\omega^n+\omega^{2 n}\right) \\ = \begin{cases}\left(2^2\right)^{\frac{n}{3}} \text { if } n \equiv 0(\bmod 3) \\ \left(2^2\right)^{\left[\frac{n}{3}\right]} \cdot(-2 \omega) \text { if } n \equiv 1(\bmod 3) \\ \left(2^2\right)^{\left[\frac{n}{3}\right]}(-2 \omega)\left(-2 \omega^2\right) \text { if } n \equiv 2(\bmod 3)\end{cases}$

Equation of a line in the complex plane is $\bar{\alpha} \bar{z}+\alpha z+\beta=0$

where $0 \neq \alpha \in \mathbb{C}, \beta \in \mathbb{R}$, and $z=x+i y \in \mathbb{C}$.

Proof: Ax+By+C = 0

where, A,B,C $\in {\R}$ & $A^2+B^2 \neq 0$

Let $z=x+i y$

$x=\frac{z+\bar{z}}{2} \quad y=\frac{z-\bar{z}}{2 i} \\ \Rightarrow A\left(\frac{z+\bar{z}}{2}\right)+B\left(\frac{z-\bar{z}}{2 i}\right)+C=0 \\ \Rightarrow \quad \bar{z}\left(\frac{A+i B}{2}\right)+z\left(\frac{A-i B}{2}\right)+C=0$

put $\alpha=\frac{A-i B}{2}$

${\bar{\alpha} \bar{z}+\alpha z+c=0} \rightarrow \{1\}$

$\because A^2+B^2 \neq 0, \quad|\alpha| \neq 0, \quad \alpha \neq 0 \\ \Rightarrow \quad 0 \neq \alpha \in \mathbb{C}, \quad c \in \mathbb{R}$

slope of $A x+B y+C=0: \quad m=-\frac{A}{B} \quad B \neq 0$

slope of (1) will be

$A=\alpha+\bar{\alpha} \\ B=\bar{\alpha}-\alpha \\ m=i \frac{(\alpha+\bar{\alpha})}{\alpha-\bar{\alpha}}$

Consider the lines $d_1$ and $d_2$ with equations $\bar{\alpha}_1 \bar{z}+\alpha_1 z+\beta_1=0$ and $\bar{\alpha}_2 \bar{z}+\alpha_2 z+\beta_2=0$ respectively. Then show that the lines $d_1$ and $d_2$ are:

(i) parallel iff $\frac{\bar{\alpha}_1}{\alpha_1}=\frac{\bar{\alpha}_2}{\alpha_2}$

(ii) Perpendicular iff $\frac{\overline{\alpha_1}}{\alpha_2}+\frac{\overline{\alpha_2}}{\alpha_2}=0$

$|z|=1 \\ z=1 \cdot \operatorname{cis} \theta \text {, } \\ 0 \leqslant \theta<2 \pi \\ \rightarrow|z|=r \\ z=r \operatorname{cis} \theta, 0 \leq \theta<2 \pi$

$\rightarrow$ circle with center as origin : radius $=r$

General equation of a circle:

$z-z_0 =r \operatorname{cis} \theta, 0 \leqslant \theta<2 \pi \\ \left|z-z_0\right| =r$

$(*) \leftarrow\left|z-z_0\right|^2=r^2 \\ z_0=x_0+ i y_0, z=x+i y \\ \left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$

$(*) \rightarrow\left(z-z_0\right)\left(\bar{z}-\bar{z}_0\right)=r^2 \\ z \bar{z}-z \bar{z}_0-\bar{z} z_0+\left|z_0\right|^2-r^2=0 \\ z \bar{z}-z \bar{z}_0-\bar{z} z_0+c=0 \rightarrow (I)$

$\text {(I) describes a circle if } c=\left|z_0\right|^2-r^2$

$\Rightarrow r^2=\left|z_0\right|^2-c>0$

i.e, $c<\left|\mathrm{z}_0\right|^2$

Prove that $|z-\alpha|^2+|z-\beta|^2=k$ represents a circle iff $|\alpha-\beta|^2<2 k$

$|z-\alpha|^2+|z-\beta|^2=k$

$(z-\alpha)(\bar{z}-\bar{\alpha})+(z-\beta)(\bar{z}-\bar{\beta})=k \\ 2 z \bar{z}-z \bar{\alpha}-\bar{z} \alpha+|\alpha|^2 -z \bar{\beta}-\bar{z} \beta+|\beta|^2=k \\ 2 z \bar{z}-z(\bar{\alpha}+\bar{\beta})-\bar{z}(\alpha+\beta)+|\alpha|^2+|\beta|^2-k=0 \\ z \bar{z}-z\left(\frac{\bar{\alpha}+\bar{\beta}}{2}\right)-\bar{z}\left(\frac{\alpha+\beta}{2}\right)+c=0 \leftarrow (*)$

$(*)$ is circle $\Leftrightarrow$ $c < \left|\frac{\alpha + \beta}{2}\right|^2$

$c < \left|\frac{\alpha + \beta}{2}\right|^2$

$\frac{\left|\alpha \right|^2+\left|\beta \right|^2-K}{2}< \frac{1}{4} \{\left|\alpha \right|^2+\left|\beta \right|^2+\alpha \bar {\beta}+ \bar{\alpha}\beta\}$

$-2k< - |\alpha|^2-|\beta|^2 + \alpha \bar{\beta} + \bar{\alpha}\beta$

$|\alpha - \beta|^2<2k$

Prove that

$|z-\alpha|=k|z-\beta|$

represents a circle, where $\alpha, \beta \in \mathbb C$, $k>0, k \neq 1$

Let complex numbers $\alpha$ and $\frac{1}{\alpha}$ lies on circles $\left(x-x_0\right)^2+\left(y-y_0\right)^2= r^2$ and $\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2$, respectively, If $z_0=x_0+i y_0$ satisfies the equation $2|z_0|^2=r^2+2$, then $|\alpha|$ is equal to

(1) $\frac{1}{\sqrt{2}}$ $\quad(2) \frac{1}{2}$ $\quad (3)\frac{1}{\sqrt{7}}$ $\quad(4)\frac{1}{3}$

Thank you