Complex Numbers

Prove that $\left(1 + i\right)^n + \left(1 - i\right)^n = 2 ^{\frac{n + 2}{2}} \cos \left(\dfrac{n \pi}{4}\right)$, $\;$ $n \in N$

Let $\;$ $\left(1 + i\right) = r \left(\cos \theta + i \sin \theta\right)$ $\;\;\; \cdots \; (1)$

Equating the real and imaginary parts separately, we have

$r \; \cos \theta = 1$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $r \; \sin \theta = 1$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $r = \sqrt{\left(1\right)^2 + \left(1\right)^2} = \sqrt{2}$

Substituting the value of $r$ in equations $(2a)$ and $(2b)$, we have

$\cos \theta = \dfrac{1}{\sqrt{2}}$, $\;\;$ $\sin \theta = \dfrac{1}{\sqrt{2}}$ $\implies$ $\theta = \dfrac{\pi}{4}$

Substituting the values of $r$ and $\theta$ in equation $(1)$ we have,

$\left(1 + i\right) = \sqrt{2} \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]$

$\begin{aligned} \therefore \; \left(1 + i\right)^n & = \left\{\sqrt{2} \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]\right\}^{n} \\\\ & = \left(\sqrt{2}\right)^n \left[\cos \left(\dfrac{\pi}{4}\right) + i \sin \left(\dfrac{\pi}{4}\right)\right]^n \\\\ & = 2^{\frac{n}{2}} \left[\cos \left(\dfrac{n \pi}{4}\right) + i \sin \left(\dfrac{n \pi}{4}\right)\right] \;\;\; \cdots \; (3a) \end{aligned}$

Replacing $+i$ with $-i$ in equation $(3a)$, we have

$\left(1 - i\right)^n = 2^{\frac{n}{2}} \left[\cos \left(\dfrac{n \pi}{4}\right) - i \sin \left(\dfrac{n \pi}{4}\right)\right]$ $\;\;\; \cdots \; (3b)$

Adding equations $(3a)$ and $(3b)$, we have

$\left(1 + i\right)^n + \left(1 - i\right)^n = 2^{\frac{n}{2}} \left[2 \cos \left(\dfrac{n \pi}{4}\right)\right]$

i.e. $\;$ $\left(1 + i\right)^n + \left(1 - i\right)^n = 2 ^{\frac{n + 2}{2}} \cos \left(\dfrac{n \pi}{4}\right)$

Hence proved.

Complex Numbers

If $\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma$, prove that

1. $\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma = 3 \cos \left(\alpha + \beta + \gamma\right)$


2. $\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma = 3 \sin \left(\alpha + \beta + \gamma\right)$

Given $\;\;$ $\cos \alpha + \cos \beta + \cos \gamma = 0 = \sin \alpha + \sin \beta + \sin \gamma$ $\;\;\; \cdots \; (1)$

Let

$a = \cos \alpha + i \sin \alpha = e^{i \alpha}$ $\;\;\; \cdots \; (2a)$

$b = \cos \beta + i \sin \beta = e^{i \beta}$ $\;\;\; \cdots \; (2b)$

$c = \cos \gamma + i \sin \gamma = e^{i \gamma}$ $\;\;\; \cdots \; (2c)$

$\therefore \; a + b + c = \left(\cos \alpha + \cos \beta + \cos \gamma\right) + i \left(\sin \alpha + \sin \beta + \sin \gamma\right)$

i.e. $\;$ $a + b + c = 0$ $\;\;$ [from equation $(1)$] $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ We have from equation $(3)$, $\;$ $a^3 + b^3 + c^3 = 3 abc$ $\;\;\; \cdots \; (4)$

[Note:

$a + b + c = 0$ $\implies$ $a + b = -c$ $\;\;\; \cdots \; (A)$

$\therefore$ $\;$ $\left(a + b\right)^3 = \left(-c\right)^3$

i.e. $\;$ $a^3 + b^3 + 3ab \left(a + b\right) = - c^3$

i.e. $\;$ $a^3 + b^3 + c^3 = - 3 ab \left(a + b\right)$

i.e. $\;$ $a^3 + b^3 + c^3 = 3abc$ $\;$ (from equation $A$) ]

From equation $(2a)$,

$a^3 = \left(\cos \alpha + i \sin \alpha\right)^3 = \cos 3 \alpha + i \sin 3 \alpha$ $\;\;\; \cdots \; (5a)$

[By De Moivre's theorm: $\;$ $\left(\cos \theta + i \sin \theta \right)^n = \cos n \theta + i \sin n \theta$]

From equation $(2b)$,

$b^3 = \left(\cos \beta + i \sin \beta\right)^3 = \cos 3 \beta + i \sin 3 \beta$ $\;\;\; \cdots \; (5b)$

From equation $(2c)$,

$c^3 = \left(\cos \gamma + i \sin \gamma\right)^3 = \cos 3 \gamma + i \sin 3 \gamma$ $\;\;\; \cdots \; (5c)$

Adding equations $(5a)$, $(5b)$ and $(5c)$ we get,

$a^3 + b^3 + c^3 = \left(\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma\right)$

$\hspace{3cm}$ $+ i \left(\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma\right)$ $\;\;\; \cdots \; (6a)$

We have from equations $(2a)$, $(2b)$ and $(2c)$,

$\begin{aligned} 3abc & = 3 e^{i \alpha} \cdot e^{i \beta} \cdot e^{i \gamma} \\\\ & = 3 e^{i \left(\alpha + \beta + \gamma\right)} \\\\ & = 3 \left[\cos \left(\alpha + \beta + \gamma\right) + i \sin \left(\alpha + \beta + \gamma\right)\right] \\\\ & = 3 \cos \left(\alpha + \beta + \gamma\right) + 3 i \sin \left(\alpha + \beta + \gamma\right) \;\;\; \cdots \; (6b) \end{aligned}$

$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(4)$ becomes

$\left(\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma\right) + i \left(\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma\right)$

$\hspace{4cm}$ $= 3 \cos \left(\alpha + \beta + \gamma\right) + 3 i \sin \left(\alpha + \beta + \gamma\right)$ $\;\;\; \cdots \; (7)$

1. Equating the real parts on either side of equation $(7)$ we have,
   
   $\cos 3 \alpha + \cos 3 \beta + \cos 3 \gamma = 3 \cos \left(\alpha + \beta + \gamma\right)$



2. Equating the imaginary parts on either side of equation $(7)$ we have,
   
   $\sin 3 \alpha + \sin 3 \beta + \sin 3 \gamma = 3 \sin \left(\alpha + \beta + \gamma\right)$

Complex Numbers

Simplify: $\;\;\;$ $\dfrac{\left(\cos 2 \theta - i \sin 2 \theta\right)^7 \left(\cos 3 \theta + i \sin 3 \theta\right)^{-5}}{\left(\cos 4 \theta + i \sin 4 \theta\right)^{12} \left(\cos 5 \theta - i \sin 5 \theta\right)^{-6}}$

$\begin{aligned} \dfrac{\left(\cos 2 \theta - i \sin 2 \theta\right)^7 \left(\cos 3 \theta + i \sin 3 \theta\right)^{-5}}{\left(\cos 4 \theta + i \sin 4 \theta\right)^{12} \left(\cos 5 \theta - i \sin 5 \theta\right)^{-6}} & = \dfrac{\left(e^{-i2 \theta}\right)^7 \cdot \left(e^{i 3 \theta}\right)^{-5}}{\left(e^{i 4 \theta}\right)^{12} \cdot \left(e^{-i 5 \theta}\right)^{-6}} \\\\ & = \dfrac{\left(e^{- i 14 \theta}\right) \cdot \left(e^{- i 15 \theta}\right)}{\left(e^{i 48 \theta}\right) \cdot \left(e^{i 30 \theta}\right)} \\\\ & = \dfrac{e^{- i 29 \theta}}{e^{i 78 \theta}} \\\\ & = e^{- i 107 \theta} \\\\ & = \cos \left(107 \theta\right) - i \sin \left(107 \theta\right) \end{aligned}$

Complex Numbers

Solve the equation $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = 0$ $\;$ if $\;$ $3 + i$ $\;$ is a root.

Given: $\;\;$ One root is $\left(3 + i\right)$

$\implies$ $\left(3 - i\right)$ is also a root.

Sum of roots $= 6$

Product of roots $= \left(3 + i\right) \left(3 - i\right) = 9 - i^2 = 10$

$\therefore$ $\;$ The corresponding factor is $\;$ $x^2 - 6x + 10$

$\therefore$ $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = \left(x^2 - 6x + 10\right) \left(x^2 + \lambda x + 2\right)$ $\;\;\; \cdots \; (1)$

$\lambda$ $\;$ is a real number (constant).

Equating the coefficients of the $x$ term in equation $(1)$ we have,

$- 32 = -12 + 10 \lambda$

i.e. $\;$ $10 \lambda = -20$ $\implies$ $\lambda = -2$

Substituting the value of $\lambda$ in equation $(1)$ we have,

$x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = \left(x^2 - 6x + 10\right) \left(x^2 - 2x + 2\right)$

$\therefore$ $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = 0$ $\implies$ $x^2 - 6x + 10 = 0$ $\;$ or $\;$ $x^2 - 2x + 2 = 0$

Now,

$\begin{aligned} x^2 - 2x + 2 = 0 \implies x & = \dfrac{2 \pm \sqrt{4 - 8}}{2} \\\\ & = \dfrac{2 \pm 2i}{2} \\\\ & = 1 \pm i \end{aligned}$

$\therefore$ $\;$ The roots are $\left(3 \pm i\right)$ and $\left(1 \pm i\right)$.

Complex Numbers

$P$ represents the variable complex number $z$. Find the locus of $P$, if $\;$ $arg \left(\dfrac{z - 1}{z + 3}\right) = \dfrac{\pi}{4}$

Let $z = x + iy$

Given: $\;\;\;$ $arg \left(\dfrac{z - 1}{z + 3}\right) = \dfrac{\pi}{4}$

i.e. $\;$ $arg \left(z - 1\right) - arg \left(z + 3\right) = \dfrac{\pi}{4}$

i.e. $\;$ $arg \left[\left(x - 1\right) + i y\right] - arg \left[\left(x + 3\right) + i y\right] = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left(\dfrac{y}{x - 1}\right) - \tan^{-1} \left(\dfrac{y}{x + 3}\right) = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left[\dfrac{\dfrac{y}{x - 1} - \dfrac{y}{x + 3}}{1 + \left(\dfrac{y}{x - 1}\right) \left(\dfrac{y}{x + 3}\right)} \right] = \dfrac{\pi}{4}$

i.e. $\;$ $\tan^{-1} \left[\dfrac{4y}{x^2 + 2x - 3 + y^2}\right] = \dfrac{\pi}{4}$

i.e. $\;$ $\dfrac{4y}{x^2 + y^2 + 2x - 3} = \tan \left(\dfrac{\pi}{4}\right) = 1$

i.e. $\;$ $x^2 + y^2 + 2x - 4y - 3 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of locus of $P$.

Complex Numbers

$P$ represents the variable complex number $z$. Find the locus of $P$, if $\;$ $\left|z - 5i\right| = \left|z + 5i\right|$

Let $z = x + iy$

Then, $z - 5i = x + i \left(y - 5\right)$; $\;\;\;$ $z + 5i = x + i \left(y + 5\right)$

$\left|z - 5i\right| = \sqrt{\left(x^2\right) + \left(y - 5\right)^2} = \sqrt{x^2 + y^2 - 10 y + 25}$

$\left|z - 5i\right| = \sqrt{\left(x^2\right) + \left(y + 5\right)^2} = \sqrt{x^2 + y^2 + 10 y + 25}$

$\therefore$ $\;$ $\left|z - 5i\right| = \left|z + 5i\right|$ $\implies$ $\sqrt{x^2 + y^2 - 10 y + 25} = \sqrt{x^2 + y^2 + 10 y + 25}$

i.e. $\;$ $x^2 + y^2 - 10 y + 25 = x^2 + y^2 + 10 y + 25$

i.e. $\;$ $y = 0$

$\therefore$ $\;$ The locus of point $P$ is $y = 0$ $\;$ i.e. $\;$ the $X$ axis.

Complex Numbers

$P$ represents the variable complex number $z$.Find the locus of $P$, if $\;$ $Im \left[\dfrac{2z + 1}{iz + 1}\right] = -2$

Let $P$ be the point $z = x + iy$

Then,

$\begin{aligned} \dfrac{2z + 1}{iz + 1} & = \dfrac{\left(2x + 1\right) + i \; 2y}{\left(1 - y\right) + i \; x} \\\\ & = \dfrac{\left[\left(2x + 1\right) + i \; 2y\right] \left[\left(1 - y\right) - i \;x\right]}{\left(1 -y\right)^2 - \left(i \; x\right)^2} \\\\ & = \dfrac{\left(2x + 1\right) \left(1 - y\right) - i \; x \left(2x + 1\right) + i \; 2y \left(1 - y\right) - i^2 \; 2xy}{1 -2y + y^2 - i^2 \; x^2} \\\\ & = \dfrac{\left(2x + 1\right) \left(1 - y\right) + 2xy}{x^2 + y^2 - 2y + 1} + i \;\dfrac{2y \left(1 - y\right) - x \left(2x + 1\right)}{x^2 + y^2 - 2y + 1} \end{aligned}$

$\therefore$ $\;$ $Im \left[\dfrac{2z + 1}{iz + 1}\right] = \dfrac{2y - 2 y^2 -2 x^2 - x}{x^2 + y^2 - 2y + 1}$

Given $\;\;$ $\dfrac{2y - 2 y^2 -2 x^2 - x}{x^2 + y^2 - 2y + 1} = -2$

i.e. $\;$ $2 y - 2 y^2 - 2 x^2 - x = - 2 x^2 - 2 y^2 + 4 y - 2$

i.e. $\;$ $x + 2y - 2 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is the required equation of locus of $P$.