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Maths Solve the Following Question.(2 Marks)

Complex Numbers · Maharashtra Board STD 11 Science (MSBSHSE - English Medium). 72 questions.

Questions · Page 1 of 2

Solve the Following Question.(2 Marks)

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Question 12 Marks
If $\omega$ is the cube root of unity, then find the value of $\left(\frac{-1+i \sqrt{3}}{2}\right)^{18}+\left(\frac{-1-i \sqrt{3}}{2}\right)^{18}$
Answer
If ω is the complex cube root of unity, then
$\omega^3=1, \omega=\frac{-1+i \sqrt{3}}{2}$ and $\omega^2=\left(\frac{-1-i \sqrt{3}}{2}\right)^2$
Consider, $\left(\frac{-1+ i \sqrt{3}}{2}\right)^{18}+\left(\frac{-1- i \sqrt{3}}{2}\right)^{18}$
Given Expression $=\omega^{18}+\left(\omega^2\right)^{18}$
$=\omega^{18}+\omega^{36}$
$=\left(\omega^3\right)^6+\left(\omega^3\right)^{12}$
$=(1)^6+(1)^{12}$
$=2$
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Question 132 Marks
Simplify the following and express in the form a + ib:
$\frac{5+7 i}{4+3 i}+\frac{5+7 i}{4-3 i}$
Answer
$\frac{5+7 i}{4+3 i}+\frac{5+7 i}{4-3 i}$
$=(5+7 i)\left[\frac{1}{4+3 i}+\frac{1}{4-3 i}\right]$
$=(5+7 i)\left[\frac{4-3 i+4+3 i}{(4+3 i)(4-3 i)}\right]$
$=(5+7 i)\left[\frac{8}{16-9 i^2}\right]$
$=(5+7 i) \cdot\left[\frac{8}{16-9(-1)}\right]$
$=\frac{8(5+7 i)}{25}=\frac{40+56 i}{25}$
$=\frac{40}{25}+\frac{56}{25} i$
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Question 142 Marks
Simplify the following and express in the form $a + ib:$
$\frac{3 i^5+2 i^7+i^9}{i^6+2 i^8+3 i^{18}}$
Answer
$ \frac{3 i^5+2 i^7+i^9}{i^6+2 i^8+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}$
$= \frac{3(1) \cdot i+2(1)(-i)+(1)^2 \cdot i}{(1)(-1)+2(1)^2+3(-1)^9}$
$\quad \ldots\left[\because i^2=-1, i^3=-i, i^4=1\right]$
$= \frac{3 i-2 i+i}{-1+2-3}$
$=\frac{2 i }{-2}$
$=- i$
$=0- i$
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Question 152 Marks
Simplify the following and express in the form a + ib:
$\frac{\sqrt{5}+\sqrt{3 i}}{\sqrt{5}-\sqrt{3} i}$
Answer
$\frac{\sqrt{5}+\sqrt{3} i}{\sqrt{5}-\sqrt{3} i}$
$=\frac{(\sqrt{5}+\sqrt{3} i)(\sqrt{5}+\sqrt{3} i)}{(\sqrt{5}-\sqrt{3} i)(\sqrt{5}+\sqrt{3} i)}$
$=\frac{5+2 \sqrt{15} i+3 i^2}{5-3 i^2}$
$=\frac{5+2 \sqrt{15} i+3(-1)}{5-3(-1)}$
$=\frac{2+2 \sqrt{15} i}{8}=\frac{1+\sqrt{15} i}{4} \quad \ldots\left[\because i^2=-1\right]$
$=\frac{1}{4}+\frac{\sqrt{15} i }{4}$
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Question 162 Marks
Simplify the following and express in the form $a + ib:$
$\left(1+\frac{2}{i}\right)\left(3+\frac{4}{i}\right)(5+i)^{-1}$
Answer
$\left(1+\frac{2}{i}\right)\left(3+\frac{4}{ i }\right)(5+ i )^{-1}$
$=\frac{( i +2)}{ i } \cdot \frac{(3 i +4)}{ i } \cdot \frac{1}{5+ i }$
$=\frac{3 i ^2+4 i +6 i +8}{ i ^2(5+ i )}=\frac{-3+10 i +8}{-1(5+ i )} \cdots\left[\because i ^2=-1\right]$
$=\frac{(5+10 i )}{-(5+ i )}=\frac{(5+10 i )(5- i )}{-(5+ i )(5- i )}$
$=\frac{25-5 i +50 i -10 i ^2}{-\left(25- i ^2\right)}$
$=\frac{25+45 i -10(-1)}{-[25-(-1)]}=\frac{35+45 i }{-26}$
$=\frac{-35}{26}-\frac{45}{26} i$
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Question 172 Marks
Simplify the following and express in the form $a + ib:$
$(1+3 i)^2(3+i)$
Answer
$(1+3 i)^2(3+i)$
$=\left(1+6 i+9 i^2\right)(3+i)$
$=(1+6 i-9)(3+i) \ldots \ldots . .\left[\because i^2=-1\right]$
$=(-8+6 i)(3+i)$
$=-24-8 i+18 i+6 i^2$
$=-24+10 i+6(-1)$
$=-24+10 i-6$
$=-30+10 i$
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Question 182 Marks
Use De Moivre’s theorem and simplify the following:

$\frac{\cos 5 \theta+i \sin 5 \theta}{(\cos 3 \theta-i \sin 3 \theta)^2}$

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Question 192 Marks
Use De Moivre’s theorem and simplify the following:

$\frac{(\cos 2 \theta+i \sin 2 \theta)^7}{(\cos 4 \theta+i \sin 4 \theta)^3}$

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Question 202 Marks
Find the equation in cartesian coordinates of the locus of $z$ if : $\frac{|z+3 i|}{|z-6 i|}=1$
Answer
Let $z = x + iy$
$\left|\frac{z+3 i}{z-6 i}\right|=1$
$\frac{|x+i y+3 i|}{|x+i y-6 i|}=1$
$\frac{|x+(y+3) i|}{|x+(y-6) i|}=1$
$\frac{\sqrt{x^2+(y+3)^2}}{\sqrt{x^2+(y-6)^2}}=1$
$\frac{x^2+(y+3)^2}{x^2+(y-6)^2}=1$
$x^2+(y+3)^2=x^2+(y-6)^2$
$y^2+6 y+9=y^2-12 y+36$
$18 y-27=0$
$2 y-3=0$
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Question 212 Marks
Find the equation in cartesian coordinates of the locus of $z$ if $: |z – 2 – 2i | = |z + 2 + 2i|$
Answer
Let $z = x + iy$
$|z-2-2 i|=|z+2+2 i|$
$|x+i y-2-2 i|=|x+i y+2+2 i|$
$|(x-2)+i(y-2)|=|(x+2)+i(y+2)|$
$\sqrt{(x-2)^2+(y-2)^2}=\sqrt{(x+2)^2+(y+2)^2}$
$(x-2)^2+(y-2)^2=(x+2)^2+(y+2)^2$
$x^2-4 x+4+y^2-4 y+4=x^2+4 x+4+y^2+4 y+4$
$-4 x-4 y=4 x+4 y$
$8 x+8 y=0$
$x+y=0$
$y=-x$
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Question 222 Marks
Find the equation in cartesian coordinates of the locus of $z$ if $: |z + 8| = |z – 4|$
Answer
Let $z = x + iy$
$|z+8|=|z-4|$
$|x+i y+8|=|x+i y-4|$
$|(x+8)+i y|=|(x-4)+i y|$
$\sqrt{(x+8)^2+y^2}=\sqrt{(x-4)^2+y^2}$
$(x+8)^2+y^2=(x-4)^2+y^2$
$x^2+16 x+64+y^2=x^2-8 x+16+y^2$
$16 x+64=-8 x+16$
$24 x+48=0$
$\therefore x+2=0$
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Question 232 Marks
If ω is the complex cube root of unity, show that

$(a+b)^2+\left(a \omega+b \omega^2\right)^2+\left(a \omega^2+b \omega\right)^2=6 a b$

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Question 252 Marks
If ω is the complex cube root of unity, show that

$(a+b)+\left(a \omega+b \omega^2\right)+\left(a \omega^2+b \omega\right)=0$

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Question 272 Marks
If ω is the complex cube root of unity, show that

$\left(3+3 \omega+5 \omega^2\right)^6-\left(2+6 \omega+2 \omega^2\right)^3=0$

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Question 282 Marks
If ω is the complex cube root of unity, show that

$\left(2+\omega+\omega^2\right)^3-\left(1-3 \omega+\omega^2\right)^3=65$

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Question 482 Marks
Show that $\left(\frac{\sqrt{7}+i \sqrt{3}}{\sqrt{7}-i \sqrt{3}}+\frac{\sqrt{7}-i \sqrt{3}}{\sqrt{7}+i \sqrt{3}}\right)$ is real.
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