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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 2 of 2

Solve the Following Question.(2 Marks)

Question 562 Marks
Express the following in the form of $a + ib, a, b \in R, i = \sqrt{-1}.$ State the values of $a$ and $b:$
$\frac{4 i^8-3 i^9+3}{3 i^{11}-4 i^{10}-2}$
Answer
$ \frac{4 i^8-3 i^9+3}{3 i^{11}-4 i^{10}-2}=\frac{4\left(i^4\right)^2-3\left(i^4\right)^2 \cdot i+3}{3\left(i^4\right)^2 \cdot i^3-4\left(i^4\right)^2 \cdot i^2-2}$
$\because i^2=-1, i^3=-i \text { and } i^4=1$
$\therefore \frac{4 i^8-3 i^9+3}{3 i^{11}-4 i^{10}-2}=\frac{4(1)^2-3(1)^2 \cdot i+3}{3(1)^2(-i)-4(1)^2(-1)-2}$
$=\frac{4-3 i+3}{-3 i+4-2}$
$=\frac{7-3 i}{2-3 i}$
$=\frac{(7-3 i)(2+3 i)}{(2-3 i)(2+3 i)}$
$=\frac{14+21 i-6 i-9 i^2}{4-9 i^2}$
$=\frac{14+15 i-9(-1)}{4-9(-1)}$
$=\frac{23+15 i}{13}$
$\therefore \quad \frac{4 i^8-3 i^9+3}{3 i^{11}-4 i^{10}-2}=\frac{23}{13}+\frac{15}{13} i$
$\therefore \quad a=\frac{23}{13} \text { and } b=\frac{15}{13}$
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Question 572 Marks
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b: (2 + 3i)(2 – 3i)
Answer
$(2+3 i)(2-3 i)$
$=4-9 i^2$
$=4-9(-1) \ldots\left[\because i^2=-1\right]$
$=4+9$
$=13$
$\therefore(2+3 i)(2-3 i)=13+0 i$
$\therefore a=13 \text { and } b=0$
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Question 582 Marks
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b: (-√5 + 2√-4 ) + (1 – √-9 ) + (2 + 3i)(2 – 3i)
Answer
(-√5 + 2√-4) + (1 – √-9) + (2 + 3i)(2 – 3i)
$=(-\sqrt{ } 5+2 \sqrt{ } 4 \cdot \sqrt{ }-1)+(1-\sqrt{ } 9 \cdot \sqrt{ }-1)+4-9 i^2$
$=[-\sqrt{ } 5+2(2) i]+(1-3 i)+4-9 i^2$
$=-\sqrt{ } 5+4 i+1-3 i+4-9(-1) \ldots \ldots .\left[\because i^2=-1\right]$
= (14 – √5) + i
∴ a = 14 – √5 and b = 1
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Question 592 Marks
Express the following in the form of $a + ib, a, b ∈ R, i = \sqrt{-1}.$ State the values of $a$ and $b:$
$\frac{2+\sqrt{-3}}{4+\sqrt{-3}}$
Answer
$\frac{2+\sqrt{-3}}{4+\sqrt{-3}} & =\frac{2+\sqrt{3} i}{4+\sqrt{3} i}$
$=\frac{(2+\sqrt{3} i)(4-\sqrt{3} i)}{(4+\sqrt{3} i)(4-\sqrt{3} i)}$
$ =\frac{8-2 \sqrt{3} i+4 \sqrt{3} i-3 i^2}{16-3 i^2}$
$=\frac{8+2 \sqrt{3} i-3(-1)}{16-3(-1)} \quad \cdots\left[\because i^2=-1\right]$
$ =\frac{11+2 \sqrt{3} i}{19 }$
$\therefore \quad \frac{2+\sqrt{-3}}{4+\sqrt{-3}}=\frac{11}{19}+\frac{2 \sqrt{3}}{19} i$
$\therefore \quad a =\frac{11}{19} \text { and } b=\frac{2 \sqrt{3}}{19}$
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Question 602 Marks
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b:

$(1+i)^{-3}$

Answer
$\begin{aligned}(1+i)^{-3} & =\frac{1}{(1+i)^3} \\ & =\frac{1}{1+3 i+3 i^2+i^3} \\ & \left.=\frac{1}{1+3 i-3-i} \quad \ldots\left[\because{ }^3+b\right)^3=a^3+3 a^2 b+3 a b^2+b^3\right] \\ & =\frac{1}{-2+2 i}\end{aligned}$

$\begin{aligned} & =\frac{1}{2(-1+i)} \\ & =\frac{1}{2} \times \frac{-1-i}{(-1+i)(-1-i)} \\ & =\frac{1}{2} \times \frac{(-1-i)}{1-i^2} \\ & =\frac{-(1+i)}{2[1-(-1)]} \quad \ldots\left[\because i^2=-1\right] \\ & =\frac{-1}{4}(1+i) \\ \therefore \quad(1+i)^{-3} & =\frac{-1}{4}-\frac{1}{4} i \\ \therefore \quad a=b & =\frac{-1}{4}\end{aligned}$

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Question 612 Marks
Express the following in the form of $a + ib, a, b \in R, i = \sqrt{-1}.$ State the values of $a$ and $b:$
$\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}$
Answer
$ \frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}$
$=  \frac{(3+2 i)(2+5 i)+(2-5 i)(3-2 i)}{(2-5 i)(2+5 i)}$
$=  \frac{6+15 i+4 i+10 i^2+6-4 i-15 i+10 i^2}{4-25 i^2}$
$=  \frac{12+20 i^2}{4-25 i^2} \quad \ldots\left[\because i^2=-1\right]$
$=  \frac{12+20(-1)}{4-25(-1)}$
$=\frac{-8}{29}$
$\therefore \quad \frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}=\frac{-8}{29}+0 i$
$\therefore \quad a=\frac{-8}{29} \text { and } b=0$
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Question 622 Marks
Express the following in the form of $a + ib, a, b \in R, i = \sqrt{-1}.$ State the values of $a$ and $b:$
$\left(\frac{1+i}{1-1}\right)^2$
Answer
$\left(\frac{1+ i }{1- i }\right)^2  =\frac{1+2 i + i ^2}{1-2 i + i ^2}$
$ =\frac{1+2 i -1}{1-2 i -1} \quad \ldots\left[\because i ^2=-1\right]$
$ =\frac{2 i }{-2 i }$
$ =-1$
$\therefore \quad\left(\frac{1+i}{1-i}\right)^2=-1+0 i$
$\therefore \quad a=-1 \text { and } b=0$
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Question 632 Marks
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b:
$\frac{(2+i)}{(3-i)(1+2 i)}$
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Question 642 Marks
Express the following in the form of $a + ib, a, b \in R, i = \sqrt{-1}.$ State the values of a and b:
$\frac{i(4+3 i)}{1-i}$
Answer
$\frac{i(4+3 i)}{1-i}  =\frac{4 i+3 i^2}{1-i}$
$ =\frac{-3+4 i}{1-i} \quad \ldots\left[\because i^2=-1\right]$
$ =\frac{(-3+4 i)(1+i)}{(1-i)(1+i)}$
$ =\frac{-3-3 i+4 i+4 i^2}{1-i^2}$
$ =\frac{-3+i+4(-1)}{1-(-1)} \quad \ldots\left[\because i^2=-1\right]$
$=\frac{-7+i}{2}$
$\therefore \quad \frac{i(4+3 i)}{1-i}=\frac{-7}{2}+\frac{1}{2} i$
$\therefore \quad a=\frac{-7}{2} \text { and } b=\frac{1}{2}$
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Question 652 Marks
Express the following in the form of $a + ib, a, b \in R, i = \sqrt{-1}.$ State the values of $a$ and $b: (1 + i)(1 – i)-1$
Answer
$(1+i)(1-i)^{-1}=\frac{1+i}{1-i}$
$=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{1+2 i+i^2}{1-i^2}$
$=\frac{1+2 i-1}{1-(-1)} \quad \cdots\left[\because i^2=-1\right]$
$=\frac{2 i}{2}$
$= i$
$\therefore \quad(1+ i )(1- i )^{-1}=0+ i$
$\therefore \quad a =0 \text { and } b =1$
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Question 662 Marks
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b: (1 + 2i)(-2 + i)
Answer
$(1+2 i)(-2+i)=-2+i-4 i+2 i^2$
$=-2-3 i+2(-1) \ldots \ldots\left[\because i^2=-1\right]$
∴ (1 + 2i)(-2 + i) = -4 – 3i ∴ a = -4 and b = -3
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Question 672 Marks
Find a and b if :(a + ib) (1 + i) = 2 + i
Answer
(a + ib)(1 + i) = 2 + i a + ai + bi + bi2 = 2 + i

$a+(a+b) i+b(-1)=2+i \ldots \ldots .\left(\because i^2=-1\right)$

(a – b) + (a + b)i = 2 + i Equating real and imaginary parts, we get a – b = 2 ……(i) a + b = 1 …….(ii) Adding equations (i) and (ii), we get 2a = 3

$\therefore a=\frac{3}{2}$

Substituting $a=\frac{3}{2}$ in (ii), we get

$\frac{3}{2}+b=1$

$\therefore b=1-\frac{3}{2}=\frac{-1}{2}$

$\therefore a=\frac{3}{2}$ and $b=\frac{-1}{2}$

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Question 682 Marks
Find a and b if :
$\frac{1}{a+i b}=3-2 i$
Answer
$ \frac{1}{a+i b}=3-2 i$
$\therefore a+i b=\frac{1}{3-2 i}$
$\therefore a+i b=\frac{1}{3-2 i} \times \frac{3+2 i}{3+2 i}$
$\therefore a+i b=\frac{3+2 i}{3^2-2^2 i^2}$
$\therefore a+i b=\frac{3+2 i}{9-4(-1)} \quad \ldots\left[\because i^2=-1\right]$
$\therefore a+i b=\frac{3+2 i}{13}$
$\therefore a+i b=\frac{3}{13}+\frac{2}{13} i$
Equating real and imaginary parts, we get
$\therefore a =\frac{3}{13}$ and $b =\frac{2}{13}$
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Question 692 Marks
Find a and b if $:abi = 3a – b + 12i$
Answer
$abi = 3a – b + 12i $
$\therefore 0 + abi = (3a – b) + 12i$
 Equating real and imaginary parts,
we get $3a – b = 0 $
$\therefore 3a = b …..(i)$ and $ab = 12$
$\therefore b=\frac{12}{a}$
Substituting $b=\frac{12}{a}$ in (i), we get
$3 a=\frac{12}{a}$
$3 a^2=12$
$a^2=4$
$a= \pm 2$
When $a=2, b=\frac{12}{a}=\frac{12}{2}=6$
When $a=-2, b=\frac{12}{a}=\frac{12}{-2}=-6$
$\therefore a = 2$ and $b = 6$ or $a = -2$ and $b = -6$
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Question 702 Marks
Find a and b if :(a + b) (2 + i) = b + 1 + (10 + 2a)i
Answer
(a + b) (2 + i) = b + 1 + (10 + 2a)i 2(a + b) + (a + b)i = (b + 1) + (10 + 2a)i Equating real and imaginary parts, we get 2(a + b) = b + 1 ∴ 2a + b = 1 ……(i) and a + b = 10 + 2a -a + b = 10 …….(ii) Subtracting equation (ii) from (i), we get 3a = -9 ∴ a = -3 Substituting a = – 3 in (ii), we get -(-3) + b = 10 ∴ b = 7 ∴ a = -3 and b = 7
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Question 712 Marks
Find a and b if :(a – b) + (a + b)i = a + 5i
Answer
(a – b) + (a + b)i = a + 5i Equating real and imaginary parts, we get a – b = a ……(i) a + b = 5 ……(ii) From (i), b = 0 Substituting b = 0 in (ii), we get a + 0 = 5 ∴ a = 5 ∴ a = 5 and b = 0
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Question 722 Marks
Find a and b if : a + 2b + 2ai = 4 + 6i
Answer
a + 2b + 2ai = 4 + 6i Equating real and imaginary parts, we get a + 2b = 4 …..(i) 2a = 6 ……(ii) ∴ a = 3 Substituting, a = 3 in (i), we get 3 + 2b = 4
$\therefore b=\frac{1}{2}$
$\therefore a=3 \text { and } b=\frac{1}{2}$ Check:
For $a =3$ and $b =\frac{1}{2}$
Consider, L.H.S. = a + 2b + 2ai
$=3+2\left(\frac{1}{2}\right)+2(3) i$
= 4 + 6i = R.H.S.
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Solve the Following Question.(2 Marks) - Page 2 - Maths STD 11 Science Questions - Vidyadip