Solve the Following Question.(2 Marks)

Question 12 Marks

Solve the following equations for $x, y \in R$ : $(1-3 i ) x +(2+5 i ) y =1+ i$

Answer

$ (1-3 i) x+(2+5 i) y=7+i$
$\therefore(x+2 y)+(-3 x+5 y) i=7+i $
Equating real and imaginary parts, we get
$x+2 y=7 \ldots . . .(i)$
$\text { and }-3 x+5 y=1 . . $
Equation (i) $\times 3+$ equation (ii) gives
$11 y =22$
$\therefore y =2$
Putting $y=2$ in (i), we get
$ x+2(2)=7$
$\therefore x=3$
$\therefore x=3 \text { and } y=2$

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Question 22 Marks

Solve the following equations for $x, y \in R$ : $(4-5 i) x+(2+3 i) y=10-7 i$

Answer

$(4-5 i) x+(2+3 i) y=10-7 i$
$\therefore(4 x+2 y)+(3 y-5 x) i=10-7 i $
Equating real and imaginary parts, we get
$ 4 x+2 y=10$
$\text { i.e., } 2 x+y=5 \ldots$
$\text { and } 3 y-5 x=-7$
$\text { Equation (i) } \times 3-$
$11 x=22 $
Equation (i) $\times 3-$ equation (ii) gives
$11 x =22$
$\therefore x =2$
Putting $x=2$ in (i), we get
$2(2)+y=5$
$\therefore y=1$
$\therefore x=2 \text { and } y=1$

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Question 32 Marks

Simplify the following and express in the form $a+i b$ : $\frac{3 i^5+2 i^7+i^9}{i^6+2 i^8+3 i^{18}}$

Answer

$
\begin{aligned}
\frac{3 i^5+2 i^7+i^9}{i^6+2 i^6+3 i^{18}} & =\frac{3\left(i^4 \cdot i\right)+2\left(i^4 \cdot i^3\right)+\left(i^4\right)^2 \cdot i}{i^4 \cdot i^2+2\left(i^4\right)^2+3\left(i^2\right)^9} \\
& =\frac{3(1) \cdot i+2(1)(-i)+(1)^2 \cdot i}{(1)(-1)+2(1)^2+3(-1)^9} \\
& \quad \ldots\left[\because i^2=-1, i^3=-i, i^4=1\right] \\
& =\frac{3 i-2 i+i}{-1+2-3} \\
& =\frac{2 i}{-2} \\
& =-i=0-i
\end{aligned}
$

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Question 42 Marks

Simplify the following and express in the form $a+i b$ : $\frac{\sqrt{5}+\sqrt{3} i}{\sqrt{5}-\sqrt{3} i}$

Answer

$\frac{\sqrt{5}+\sqrt{3} i }{\sqrt{5}-\sqrt{3} i }$
$=\frac{(\sqrt{5}+\sqrt{3} i)(\sqrt{5}+\sqrt{3} i)}{(\sqrt{5}-\sqrt{3} i)(\sqrt{5}+\sqrt{3} i)}$
$=\frac{5+2 \sqrt{15} i +3 i ^2}{5-3 i ^2}$
$=\frac{5+2 \sqrt{15} i +3(-1)}{5-3(-1)}$
$\ldots\left[\because i ^2=-1\right]$
$=\frac{2+2 \sqrt{15} i }{8}=\frac{1+\sqrt{15} i }{4}$
$=\frac{1}{4}+\frac{\sqrt{15} i }{4}$

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Question 52 Marks

Simplify the following and express in the form $a+i b$ : $\left(1+\frac{2}{i}\right)\left(3+\frac{4}{i}\right)(5+i)^{-1}$

Answer

$ \left(1+\frac{2}{ i }\right)\left(3+\frac{4}{ i }\right)(5+ i )^{-1}$
$=\frac{( i +2)}{ i } \cdot \frac{(3 i +4)}{ i } \cdot \frac{1}{5+ i }$
$=\frac{3 i ^2+4 i +6 i +8}{ i ^2(5+ i )}=\frac{-3+10 i +8}{-1(5+ i )} \ldots\left[\because i ^2=-1\right]$
$=\frac{(5+10 i )}{-(5+ i )}=\frac{(5+10 i )(5- i )}{-(5+ i )(5- i )}$
$=\frac{25-5 i +50 i -10 i ^2}{-\left(25- i ^2\right)}$
$=\frac{25+45 i -10(-1)}{-[25-(-1)]}=\frac{35+45 i }{-26}$
$=\frac{-35}{26}-\frac{45}{26} i$

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Question 62 Marks

Simplify the following and express in the form $a+i b$ : $\frac{5+7 i}{4+3 i}+\frac{5+7 i}{4-3 i}$

Answer

$\frac{5+7 i }{4+3 i }+\frac{5+7 i }{4-3 i }$
$=(5+7 i )\left[\frac{1}{4+3 i }+\frac{1}{4-3 i }\right]$
$=(5+7 i )\left[\frac{4-3 i +4+3 i }{(4+3 i )(4-3 i )}\right]$
$=(5+7 i )\left[\frac{8}{16-9 i ^2}\right]$
$=(5+7 i ) \cdot\left[\frac{8}{16-9(-1)}\right]$
$=\frac{8(5+7 i )}{25}=\frac{40+56 i }{25}$
$=\frac{40}{25}+\frac{56}{25} i =\frac{8}{5}+\frac{56}{25} i $

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Question 72 Marks

If $\omega$ is a complex cube root of unity, then prove the following : $(a+b)+\left(a w+b \omega^2\right)+\left(a \omega^2+b w\right)=0$

Answer

$
\begin{aligned}
& \text {L.H.S. }=(a+b)+\left(a \omega+b \omega^2\right)+\left(a \omega^2+b \omega\right) \\
& =\left(a+a w+a \omega^2\right)+\left(b+b \omega+b \omega^2\right) \\
& =a\left(1+\omega+\omega^2\right)+b\left(1+\omega+\omega^2\right) \\
& =a(0)+b(0) \\
& =0 \\
& =\text { R.H.S. }
\end{aligned}
$

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Question 82 Marks

If $\omega$ is a complex cube root of unity, then prove the following : $\left(\omega^2+w-1\right)^3=-8$

Answer

$\omega$ is the complex cube root of unity.
$
\therefore \omega^3=1 \text { and } 1+\omega+\omega^2=0
$
Also, $1+\omega^2=-\omega, 1+\omega=-\omega^2$ and $\omega+\omega^2=-1$
$
\begin{aligned}
& \text {L.H.S. }=\left(\omega^2+\omega-1\right)^3 \\
& =(-1-1)^3 \\
& =(-2)^3 \\
& =-8 \\
& =\text { R.H.S. } \\
&
\end{aligned}
$

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Question 92 Marks

Find the value of $1+i^2+i^4+i^6+i^8+\ldots \ldots+i^{20}$.

Answer

$\begin{aligned} & 1+i^2+i^4+i^6+i^8+\ldots .+i^{20} \\ & =1+\left(i^2+i^4\right)+\left(i^6+i^8\right)+\left(i^{10}+i^{12}\right)+\left(i^{14}+i^{16}\right)+\left(i^{18}+i^{20}\right) \\ & =1+\left[i^2+\left(i^2\right)^2\right]+\left[\left(i^2\right)^3+\left(i^2\right)^4\right]+\left[\left(i^2\right)^5+\left(i^2\right)^6\right]+\left[\left(i^2\right)^7+\left(i^2\right)^8\right]+\left[\left(i^2\right)^9+\left(i^2\right)^{10}\right] \\ & =1+\left[-1+(-1)^2\right]+\left[(-1)^3+(-1)^4\right]+\left[(-1)^5+(-1)^6\right]+\left[(-1)^7+(-1)^8\right]+\left[(-1)^9+(-1)^{10}\right]\left[\because i^2=\right. \\ & -1] \\ & =1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+(-1+1) \\ & =1+0+0+0+0+0 \\ & =1\end{aligned}$

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Question 102 Marks

\text { Show that }(-1+\sqrt{ } 3 i)^3 \text { is a real number. }

Answer

\begin{aligned}
& (-1+\sqrt{ } 3 i)^3 \\
& =(-1)^3+3(-1)^2(\sqrt{ } 3 i)+3(-1)(\sqrt{ } 3 i)^2+(\sqrt{ } 3 i)^3\left[\because(a+b)^3=a^3+3 a^2 b+3 a b^2+b^3\right] \\
& =-1+3 \sqrt{ } 3 i-3\left(3 i^2\right)+3 \sqrt{ } 3 i^3 \\
& =-1+3 \sqrt{ } 3 i-3(-3)-3 \sqrt{ } 3 i\left[\because i^2=-1, i^3=-1\right] \\
& =-1+9 \\
& =8 \text {, which is a real number. }
\end{aligned}

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Question 112 Marks

Express the following in the form of $a+i b, a, b \in R, i=\sqrt{ }-1$. State the values of $a$ and $b$ : $\frac{(2+i)}{(3-i)(1+2 i)}$

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Question 122 Marks

Express the following in the form of $a+i b, a, b \in R, i=\sqrt{ }-1$. State the values of $a$ and $b$ : $\frac{\mathrm{i}(4+3 \mathrm{i})}{(1-\mathrm{i})}$

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Maths (commerce) Solve the Following Question.(2 Marks)

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