KC Sinha Mathematics Solution Class 11 Chapter 10 समिश्र संख्याएँ (complex number) Exercise 10.2

Exercise 10.2

Type-1

Question 1

Find the square root of the following :
(i) 7-24i
Sol :

$=4^{2}+(3 i)^{2}-2 \cdot 4 \cdot 3 i$

$=(4-3 i)^{2}$

$\sqrt{7-4i}=\pm \sqrt{(4-3 i)^{2}}$

$=\pm(4-3 i)$

(ii)

$-11-60 \sqrt{-1}$
Sol :

$=(5)^{2}+(6 i)^{2}-2 \cdot 5 \cdot 6 i$

$=(5-6 i)^{2}$

$\sqrt{-11-60 \sqrt{-1}}=\pm \sqrt{(5-6 i)^{2}}$

$=\pm(5-6 i)$

(iii) -8-6i
Sol :

$=\quad 1^{2}+(3 i)^{2}-2 \cdot 1 \cdot 3 i$

$=(1-3 i)^{2}$

$\sqrt{-8-6 i}=\pm \sqrt{(1-3i)^{2}}$

$=\pm(1-3 i)$

(iv) -5+12i
Sol :

$=2^{2}+(3 i)^{2}+2 \cdot 2 \cdot 3 i$

$=(2+3 i)^{2}$

$\sqrt{-5+12 i}=\pm \sqrt{(2+3 i)^{2}}$

$=\pm(2+3 i)$

(v)

$4-6 \sqrt{-5}$
Sol :

$=4-6 \sqrt{3} i$

$=3^{2}+(\sqrt{5} i)^{2}$

$=(3-\sqrt{5} i)^{2}$

$\sqrt{4-6 \sqrt{-5}}=\pm \sqrt{(3-\sqrt{5} i)^{2}}$

$=\pm(3-\sqrt{5} i)$

(vi)

$6 \sqrt{-2}-7$
Sol :

$=-7+6 \sqrt{2} i$

$=(\sqrt2)^{2}+(3 i)^{2}+2 \cdot \sqrt{2} \cdot 3 i$

$=(\sqrt{2}+3 i)^{2}$

$\sqrt{6 \sqrt{-2}-7}=\pm \sqrt{(\sqrt{2}+3 \sqrt{2})^{2}}$

$=\pm(\sqrt{2}+3 i)$

(vii)

$4 a b-2\left(a^{2}-b^{2}\right) i$
Sol :

$(a+b)^{2}+\left\{(a-b)i\}^{2}-2 \cdot(a+b)(a-b)i\right.$

$=[(a+b)-i(a-b)]^{2}$

$\sqrt{4 a b-2\left(a^{2}-b^{2}\right) i}=\pm \sqrt{[(a+b)-i(a-b)]^{2}}$

$=\pm[(a+b)-i(a-b)]$

(viii)

$\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}-\frac{1}{i}\left(\frac{x}{y}-\frac{y}{x}\right)-\frac{9}{4}$
Sol :

$=\left(\frac{x}{y}-\frac{y}{x}\right)^{2}+2 \frac{x}{y} \times \frac{y}{x}-\frac{1}{i}\left(\frac{x}{y}-\frac{y}{x}\right)-\frac{9}{4}$

$=\left(\frac{x}{y}-\frac{y}{x}\right)^{2}+2-\frac{9}{4}-\frac{1}{i}\left(\frac{x}{y}-\frac{y}{x}\right)$

$=\left(\frac{x}{y}-\frac{y}{x}\right)^{2}+\frac{8-9}{4}-\frac{1}{j}\left(\frac{x}{y}-\frac{y}{x}\right)$

$=\left(\frac{x}{y}-\frac{y}{x}\right)^{2}-\frac{1}{4}-\frac{1}{i}\left(\frac{x}{y}-\frac{y}{x}\right)$

$=\left(\frac{x}{y}-\frac{y}{x}\right)^{2}+\left(\frac{i}{2}\right)^{2}-2 \cdot\left(\frac{x}{y}-\frac{y}{x}\right) \cdot \frac{1}{2 i} \times \frac{i}{i}$

$=\left(\frac{x}{y}-\frac{y}{x}\right)^{2}+\left(\frac{i}{2}\right)^{2}+2 \cdot\left(\frac{x}{y}-\frac{y}{x}\right) \frac{i}{2}$

$=\left(\frac{x}{y}-\frac{y}{x}+\frac{1}{2}\right)^{2}$

$\sqrt{\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x}-\frac{1}{x}\left(\frac{x}{y}-\frac{y}{x}\right){-\frac{9}{4}}}$

$=\pm \sqrt{\left(\frac{x}{y}-\frac{y}{x}+\frac{i}{2}\right)^{2}}$

$=\pm\left(\frac{x}{y}-\frac{y}{x}+\frac{i}{2}\right)$

(ix)

$x^{2}+\frac{1}{x^{2}}+4 i\left(x-\frac{1}{x}\right)-6$
Sol :

$=\left(x-\frac{1}{x}\right)^{2}+2 \cdot x \cdot \frac{1}{x}-6+4i\left(x-\frac{1}{x}\right)$

$=\left(x-\frac{1}{x}\right)^{2}-4+4 i\left(x-\frac{1}{x}\right)$

$=\left(x-\frac{1}{x}\right)^{2}+(2 i)^{2}+2 \cdot\left(x-\frac{1}{x}\right) \cdot 2i$

$=\left(x-\frac{1}{x}+2 i\right)^{2}$

$\sqrt{x^{2}+\frac{1}{x^{2}}+4 i\left(x-\frac{1}{x}\right)-6}$

$=\pm \sqrt{\left(x-\frac{1}{x}+2i\right)^{2}}$

$=\pm\left(x-\frac{1}{x}+2 i\right)$

(x) 9+20i
Sol :
=9+2.10.i

$x^{2}+y^{2}=9$

xy=10

$y=\frac{10}{x}$

$x^{2}+\left(\frac{10}{x}\right)^{2}=9$

$x^{2}+\frac{100}{x^{2}}=9$

$\frac{x^{4}+100}{x^{2}}=9$

$x^{4}+100=9x^{2}$

$x^{4}-9 x^{2}+100=0$

$x^{2}=\frac{-(-9)\pm \sqrt{(-9)^2-4\times1\times100}}{2 \times 1}$

$=\frac{9 \pm \sqrt{81-400}}{2}$

$=\left(\sqrt{\frac{\sqrt{481}+9}{2}}\right)^{2}+\left(\frac{\sqrt{\sqrt{481}-9}}{2} i\right)^{2}+2 \cdot \sqrt{\frac{\sqrt{481}+9}{2}} \times \sqrt{\frac{\sqrt{481-9}}{2}}i$

$=\left(\sqrt{\frac{\sqrt{481}+9}{2}}+\sqrt{\frac{\sqrt{481}-9}{2}} i\right)^{2}$

$\sqrt{9+20 i}=\pm \left(\sqrt{\frac{\sqrt{481}+9}{2}}+\sqrt{\frac{\sqrt{41}-9}{2}} i\right)$

Question 2

Find

$\sqrt{2+3 \sqrt{-5}}+\sqrt{2-3 \sqrt{-5}}$
Sol :

$=\sqrt{2+3 \sqrt{5}} i+\sqrt{2-3 \sqrt{5}}i$

$=\sqrt{\frac{1}{2} \cdot 2(2+3 \sqrt{5}i)}+\sqrt{\frac{1}{2} \cdot 2(2-3 \sqrt{5} i)}$

$=\sqrt{\frac{1}{2}(4+6 \sqrt{5} i)}+\sqrt{\frac{1}{2}(4-6 \sqrt{5} i)}$

$=\sqrt{\left.\frac{1}{2}\left(3^{2}+(\sqrt{5})\right)^{2}+2 \cdot 3 \cdot \sqrt{5} i\right)}+\sqrt{\frac{1}{2}\left(3^{2}+(\sqrt{5}i)^{2}\right.-2 \cdot 3 \cdot \sqrt{5}i}$

$=\sqrt{\frac{1}{2}(3+\sqrt{5}i)^{2}}+\sqrt{\frac{1}{2}(3-\sqrt{5} i)^{2}}$

$=\pm \frac{1}{\sqrt{2}}(3+\sqrt{5} i) \pm \frac{1}{\sqrt{2}}(3-\sqrt{5} i)$

$=\pm \frac{1}{\sqrt{2}}(3+\sqrt{5}i + 3-\sqrt{5}i)$

$=\pm \frac{1}{\sqrt2} \times 6=\pm \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$=\pm \frac{6\sqrt{2}}{2}=\pm 3 \sqrt{2}$

Question 2

$\alpha=\frac{-1+\sqrt{-3}}{2}, \beta=\frac{-1-\sqrt{-3}}{2}$ prove that

$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+1=0$
Sol :
Let

$\alpha=\frac{-1+\sqrt{-3}}{2}=\omega$

$\beta=\frac{-1-\sqrt{-3}}{2}=\omega^{2}$

L.H.S

$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+1=\frac{\omega}{\omega^{2}}+\frac{\omega^2}{\omega}+1$

$=\omega^{2}+\omega+1=0$

Question 4

Note:

$\begin{array}{l} \text { (i)} 1+w+w^{2}=0 \\ \text { (ii)} 1+w=-w^{2} \\ \text { (iii) } 1+w^{2}=-\omega \\ \text { (iv) } w+w^{2}=-1 \\ \text { (v) } w^{3}=1 \\ \text { (vi) } \omega=\frac{1}{\omega^{2}} \\ \text { (vii) } \omega^{2}=\frac{1}{w} \end{array}$

If

$1, \omega, \omega^{2}$ be three cube roots of 1 , show that:
(i)

$\left(1+\omega-\omega^{2}\right)\left(1-\omega+\omega^{2}\right)=4$
Sol :
L.H.S

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

$=\left(-\omega^{2}-\omega^{2}\right)(-\omega-\omega)$

$=\left(-2 w^{2}\right)(-2 w)$

$=4 w^{3}$

=4(1) =4

(ii)

$\left(3+\omega+3 \omega^{2}\right)^{6}=64$
Sol :
L.H.S

$\left(3+\omega+3 \omega^{2}\right)^{6}$

$=\left[3+3 w^{2}+w\right]^{6}$

$=\left[3\left(1+w^{2}\right)+w\right]^{4}$

$=[3(-\omega)+\omega]^{6}$

$=(-2 w)^{6}=64 w^{6}$

$=64\left(\omega^{3}\right)^{2}=64(1)^{2}$

=64

Question 5

Evaluate:

$\sqrt{-2+2 \sqrt{-2+2 \sqrt{-2+\ldots \infty}}}$
Sol :
Let

$y=\sqrt{-2+2 \sqrt{-2+2 \sqrt{-2+\ldots \infty}}}$

$y=\sqrt{-2+2 y}$

Squaring both sides

$y^{2}=(\sqrt{-2+2 y})^{2}$

$y^{2}=-2+2 y$

$y^{2}-2 y+2=0$

a=1 , b=-2 , c=2

$y=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$y=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 2}}{2 \times{1}}$

$y=\frac{2 \pm \sqrt{4-8}}{2}$

$y=2 \frac{\pm \sqrt{-4}}{2}$

$y=\frac{2 \pm 2 i}{2}$

$=\frac{2(1 \pm i)}{-2}$

$=1 \pm i$

Question 6

Show that

$\left(\frac{\sqrt{3}+i}{2}\right)^{6}+\left(\frac{i-\sqrt{3}}{2}\right)^{6}=-2$
Sol :
L.H.S

$\left(\frac{\sqrt{3}+i}{2}\right)^{6}+\left(\frac{i-\sqrt{3}}{2}\right)^6$

$=\left[\left(\frac{\sqrt{3}+1}{2}\right)^{2}\right]^{3}+\left[\left(\frac{i-\sqrt{3}}{2}\right)^{2}\right]^{3}$

$=\left[\frac{(\sqrt3)^{2}+i^{2}+2 \cdot \sqrt3 \cdot i}{4}\right]^{3}+\left[\frac{i^{2}+(\sqrt 3)^{2}-2 \cdot i \sqrt{3}}{4}\right]^{3}$

$=\left[\frac{3-1+2 \sqrt{3}i}{4}\right]^{3}+\left[\frac{-1+3-2 \sqrt{3} i}{4}\right]^{3}$

$=\left[\frac{2+2 \sqrt{3} i}{4}\right]^{3}+\left[\frac{2-2 \sqrt{3} i}{4}\right]^{3}$

$=\left[\frac{2(1+\sqrt{3}i]}{4}\right]^{3}+\left[\frac{2(1-\sqrt{3} i)}{4}\right]^{3}$

$=\frac{1}{8}(1+\sqrt{3} i)^{3}+\frac{1}{8}(1-\sqrt{3} i)^{3}$

$=\frac{1}{8}\left[(1)^{3}+(\sqrt3 i)^{3}+3 \cdot 1^{2} \cdot \sqrt{3} i+3 \cdot 1 \cdot(\sqrt{3} i)^{2}+1^{3}-(\sqrt{3} i)^{3}-3.1^2.\sqrt3 i+3.1.(\sqrt{3}i)^2 \right]$

$=\frac{1}{8}[1+9(-1)+1+9(-1)]$

$=\frac{1}{8}[1-9+1-9]$

$=\frac{1}{8}(2-18)$

$=\dfrac{1}{8}(-16)$

Question 7

$1, \omega, \omega^{2}$ be the three cube roots of 1 then show that :

(i)

$(1+\omega)\left(1+\omega^{2}\right)\left(1+\omega^{4}\right)\left(1+\omega^{5}\right)=1$
Sol :
L.H.S

$(1+\omega)\left(1+\omega^{2}\right)\left(1+\omega^{4}\right)\left(1+\omega^{5}\right)$

$=\left(-\omega^{2}\right)(-\omega) \cdot\left(1+\omega^{3} \cdot \omega\right)\left(1+\omega^{3} \cdot \omega^{2}\right)$

$=w^{3} \cdot(1+w)\left(1+w^{2}\right)$

$=1 \times\left(-\omega^{2}\right)(-\omega)$

$=1 \times w^{3}$

=1×1

=1

(iii)

$\left(2 + \omega+\omega^{2}\right)^{3}+\left(1+\omega-\omega^{2}\right)^{8}-\left(1-3 \omega+\omega^{2}\right)^{4}=1$
Sol :
L.H.S

$\left(2 + \omega+\omega^{2}\right)^{3}+\left(1+\omega-\omega^{2}\right)^{8}-\left(1-3 \omega+\omega^{2}\right)^{4}$

$=(2-1)^{3}+\left(-\omega^{2}-\omega^{2}\right)^{8}-(-\omega-3 \omega)^{4}$

$=(1)^{3}+\left(-2 \omega^{2}\right)^{8}-(-4 \omega)^{4}$

$=1+256 w^{16}-256 w^{4}$

$=1+256\left(\omega^{3}\right)^{5} \cdot \omega-256 \omega^{3} \cdot \omega$

=1+256𝜔-256𝜔
=1

(iv)

$\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega^{4}\right)\left(1-\omega^{4}+\omega^{8}\right)\left(1-\omega^{8}+\omega^{16}\right) =16$
Sol :
L.H.S

$\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega^{4}\right)\left(1-\omega^{4}+\omega^{8}\right)\left(1-\omega^{8}+\omega^{16}\right)$

$=\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega^{3} \cdot \omega\right)\left(1-\omega^{3} \cdot \omega+\left(\omega^{3}\right)^{2} \omega^{2}\right)\left(1-\left(\omega^{3}\right)^{2} \omega^{2}\right.\left.+\left(\omega^{3}\right)^{5} \omega\right)$

$=\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega\right)\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega\right)$

$=(-\omega-\omega)\left(-\omega^{2}-\omega^{2}\right)(-\omega-\omega)\left(-\omega^{2}-\omega^{2}\right)$

$=(-2 \omega)\left(-2 \omega^{2}\right)(-2 \omega)\left(-2 \omega^{2}\right)$

$=4 \omega^{3} \times 4 \omega^{3}$

=4×1×4×1

=16

Question 8

If x=a+b ,

$y=a \omega+b \omega^{2}, z=a \omega^{2}+b \omega$ prove that

$x^{3}+y^{3}+z^{3}=3\left(a^{3}+b^{3}\right)$
Sol :

$x=a+b \Rightarrow x^{3}=(a+b)^{3}$

$x^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$ ..(i)

$y=a w+b w^{2} \Rightarrow y^{3}=\left(a w+b w^{2}\right)^{3}$

$y^{3}=(a \omega)^{3}+\left(b \omega^{2}\right)^{3}+3 \cdot(a \omega)^{2} \cdot\left(b \omega^{2}\right)+3 \cdot a \omega \cdot\left(b \omega^{2}\right)^{2}$

$y^{3}=a^{3} w^{3}+b^{3} w^{6}+3 a^{2} b w^{4}+3 a b^{2} w^{5}$

$y^{3}=a^{3}+b^{3} \cdot\left(w^{3}\right)^{2}+3 a^{2} b w^{3} \cdot w+3ab^2\omega^3\omega^2$

$y^{3}=a^{3}+b^{3}+3 a^{2} b w+3 a b^{2} w^{2}$ ..(ii)

$z=a w^{2}+b w \Rightarrow z^{3}=\left(a w^{2}+b w\right)^{3}$

$z^{3}=\left(a w^{2}\right)^{3}+(b w)^{3}+3 \cdot\left(a w^{2}\right)^{2} \cdot b w+3 \cdot\left(a w^{2}\right) \cdot(b w)^{2}$

$z^{3}=a^{3} w^{6}+b^{3} w^{3}+3 a^{2} b w^{5}+3ab^2\omega^4$

$z^{3}=a^{3}\left(w^{3}\right)^{2}+b^{3}+3 a^{2} b w^{3} \cdot w^{2}+3ab^2.\omega^3.\omega$

$z^{3}=a^{3}+b^{3}+3 a^{2} b w^{2}+3 a b^{2} w$ ..(iii)

Adding (i),(ii) and (iii)

$\left.x^{3}+y^{3}+z^{3}=3 a^{3}+3 b^{3}+3 a^{2} b(1+\omega)+w^{2}\right)+3 a b^{2}\left(1+w^{2}+\omega\right)$

$=3\left(a^{3}+b^{3}\right)+3 a^{2} b(0)+3 a b^{2} (0)$

$x^{3}+y^{3}+z^{3}=3\left(a^{3}+b^{3}\right)$

If x+y+z=0 then

$x^{3}+y^{3}+z^{3}=3 xxyz$

$x+y+z=a+b+a w+b w^{2}+a w^{2}+b w$

$=a\left(1+w+w^{2}\right)+b\left(1+w^{2}+w\right)$

=a(0)+b(0)

x+y+z=0

$x^{3}+y^{3}+z^{3}=3(a+b)\left(a w+b w^{2}\right)\left(a w^{2}+b w\right)$

$=3(a+b)\left[a^{2} \omega^{3}+a b \omega^{2}+a b \omega^{4}+b^{2} \omega^{3}\right]$

$=3(a+b)\left[a^{2}+a b w^{2}+a b w^{3} \cdot w+b^{2}\right]$

$=3(a+b)\left(a^{2}+a b w^{2}+a b w+b^{2}\right)$

$=3(a+b)\left[a^{2}+a b\left(\omega^{2}+\omega\right)+b^{2}\right]$

$x^{3}+y^{3}+z^{3}=3(a+b)\left[a^{2}+a b(-1)+b^{2}\right]$

$=3(a+b)\left(a^{2}-a b+b^{2}\right)$

$x^{3}+y^{3}+z^{3}=3\left(a^{3}+b^{3}\right)$

Question 9

$\left(a+b \omega+c \omega^{2}\right)^{2}+\left(a \omega \pm b \omega^{2}+c\right)^{2}+\left(a \omega+b+c \omega^{2}\right)^{2}=0$ prove that a=c or a+c=2b
Sol :

$a^{2}+b^{2} w^{2}+c^{2} b^{4}+2 a b w+2 b c w^{3}+2 c a w^{2}+a^{2} \omega^{2}+b^{2} \omega^{4}+c^{2}+2 a b \omega^{3}+2 b c \omega^{2}+2 c a \omega+a^{2} \omega^{2}+b^{2}+c^{2} \omega^{4}+2 a b \omega+2 b c \omega^{2}+2 c a \omega^{3}=0$

$a^{2}\left(1+b^{2}+w^{2}\right)+b^{2}\left(w^{2}+w^{4}+1\right)+c^{2}\left(w^{4}+1+w^{4}\right)+2 a b\left(\omega+\omega^{3}+\omega\right)+2 b c\left(\omega^{3}+\omega^{2}+\omega^{2}\right)+2 \operatorname{ca}\left(\omega^{2}+\omega_{3}\right)=0$

$a^{2}(1-1-w-1-w)+b^{2}\left(w^{2}+w+1\right)+c^{2}(w+1+w)+2 a b(w+1+w)+2 b c(1-1-w-1-w)+2 c a\left(w^{2}+w+1\right)=0$

$-a^{2}(1+2 w)+c^{2}(1+2 w)+2 a b(1+2 w)-2 b c(1+2 w) =0$

$\left[-a^{2}+c^{2}+2 a b-2 b c\right](1+2 w)=0$

$-a^{2}+c^{2}+2 a b-2 b c=0$

$c^{2}-a^{2}-2 b c+2 a b=0$

$(c-a)(c+a)-2 b(c-a)=0$

c-a=0
c=a

and

c+a-2b=0
c+a=2b

Question 10

Sol :
Let

$\alpha=\omega$

$\beta=\omega^{2}$

$x y z=(a+b)(a \alpha+b \beta)(a \beta+b \alpha)$

$=(a+b)\left(a w+b w^{2}\right)\left(a w^{2}+b w\right)$

$=(a+b)\left(a^{2} w^{3}+a b w^{2}+a b w^{4}+b^{2} w^{3}\right]$

$=(a+b)\left[a^{2}+a b w^{2}+a b w+b^{2}\right]$

$x y z=(a+b)\left[a^{2}+a b\left(w^{2}+w\right)+b^{2}\right]$

$x y 2=(a+b)\left[a^{2}+a b(-1)+b^{2}\right]$

$x y z=(a+b)\left(a^{2}-a b+b^{2}\right)$

$=a^{3}+b^{3}$

Question 11

$\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{2 n} x^{2 n}$ then prove that

$a_{0}+a_{3}+a_{6}+\ldots=3^{n-1}$
Sol :

$\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots,+a_{2 n} x^{2 n}$

x=1

$\left(1+1+1^{2}\right)^{n}=a_{0}+a_{1}(1)+a_{2}(r)^{2}+a_{3}(1)^{3}+\ldots+a_{2 n}(1)^{2n}$

$3^{n}=a_{0}+a_{1}+a_{2}+a_{3}+\cdots-\cdots+a_{2 n}$ ..(i)

$x=\omega$ ,

$\left(1+\omega+\omega^{2}\right)^{n}=a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3} \omega^{3}+\ldots-+a_{2 n} \omega^{2 n}$

$0=a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3}+-\quad+a_{2 n} w^{2 n}$ ..(ii)

$x=\omega^{2}$

$\left(1+w^{2}+w^{4}\right)^{n}=a_{0}+a_{1} w^{2}+a_{2} w^{4}+a_{3} w^{6}+--+ a_{2 n} w^{4 n}$

$\left(1+w^{2}+w\right)^{n}=a_{0}+a_{1} w^{2}+a_{2} w+a_{3}\left(w^{3}\right)^{2}+--+a_{2 n} w^{4n}$

$\theta=a_{0}+a_{1} \omega^{2}+a_{2} \omega+a_{3}+\ldots+a_{2 n} \omega^{4n}$ ..(iii)

adding (i) , (ii) and (iii)

$3^{n}=3 a_{0}+a_{1}\left(1+\omega+\omega^{2}\right)+a_{2}\left(1+\omega^{2}+\omega\right)+3 a_{3}+3 a_{6}+.....$

$3^{n}=3 a_{0}+a_{1}(0)+a_{2}(0)+3 a_{3}+3 a_{6}+....$

$3^{n}=3 a+3 a_{3}+3 a_{6}+\dots$

$3^{n}=3\left(a_{0}+a_{3}+a_{6}+\dots\right)$

$\frac{3^{n}}{3^{1}}=a_{0}+a_{3}+a_{6}+\cdots$

$3^{n-1}=a_{0}+a_{3}+a_{6}+\dots$