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Complex Numbers and Quadratic Equations

Madhya Pradesh Board · Class 11 · Mathematics

NCERT Solutions for Complex Numbers and Quadratic Equations — Madhya Pradesh Board Class 11 Mathematics.

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28 Questions Solved · 2 Sections

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EXERCISE 4.1

1(5i)(−35i)(5i)\left(-\frac{3}{5}i\right)Show solution

(5i)(−35i)=5⋅(−35)i2=−3(−1)=3(5i)\left(-\frac{3}{5}i\right)=5\cdot\left(-\frac35\right)i^2=-3(-1)=3.

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2i9+i19i^9 + i^{19}Show solution

Use the powers of ii:

i9=i4⋅2+1=i i^9=i^{4\cdot2+1}=i and i19=i4⋅4+3=−i i^{19}=i^{4\cdot4+3}=-i .

So,

i9+i19=i+(−i)=0 i^9+i^{19}=i+(-i)=0 .

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3i−39i^{-39}Show solution

i−39=1i39 i^{-39}=\frac{1}{i^{39}} .
Now 39=4⋅9+339=4\cdot9+3, so i39=i3=−ii^{39}=i^3=-i.
Hence,

i−39=1−i=i i^{-39}=\frac{1}{-i}=i .

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43(7+i7)+i(7+i7)Show solution

3(7+7i)+i(7+7i)=21+21i+7i+7i23(7+7i)+i(7+7i)=21+21i+7i+7i^2.

Since i2=−1i^2=-1,

=21+28i−7=14+28i=21+28i-7=14+28i.

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5(1 - i) - (-1 + i6)Show solution

(1−i)−(−1+6i)=1−i+1−6i=2−7i(1-i)-(-1+6i)=1-i+1-6i=2-7i.

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6(15+i25)−(4+i52)\left(\frac{1}{5} + i\frac{2}{5}\right) - \left(4 + i\frac{5}{2}\right)Show solution

Subtract real and imaginary parts separately:

Real part: 15−4=15−205=−195\frac15-4=\frac15-\frac{20}{5}=-\frac{19}{5}

Imaginary part: 25−52=410−2510=−2110\frac{2}{5}-\frac{5}{2}=\frac{4}{10}-\frac{25}{10}=-\frac{21}{10}

So the result is −195−2110i-\frac{19}{5}-\frac{21}{10}i.

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7[(13+i73)+(4+i13)]−(−43+i)\left[\left(\frac{1}{3} + i\frac{7}{3}\right) + \left(4 + i\frac{1}{3}\right)\right] - \left(-\frac{4}{3} + i\right)Show solution

First add inside the bracket:

Real part: 13+4=13+123=133\frac13+4=\frac13+\frac{12}{3}=\frac{13}{3}

Imaginary part: 73+13=83\frac73+\frac13=\frac83

So bracket becomes 133+83i\frac{13}{3}+\frac83 i.

Now subtract (−43+i)\left(-\frac43+i\right):

Real part: 133−(−43)=173\frac{13}{3}-\left(-\frac43\right)=\frac{17}{3}

Imaginary part: 83−1=53\frac83-1=\frac53

So the result is 173+53i\frac{17}{3}+\frac53 i.

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8(1−i)4(1 - i)^{4}Show solution

Using binomial expansion,

(1−i)2=1−2i+i2=1−2i−1=−2i(1-i)^2=1-2i+i^2=1-2i-1=-2i.

Then

(1−i)4=[(1−i)2]2=(−2i)2=4i2=−4(1-i)^4=[(1-i)^2]^2=(-2i)^2=4i^2=-4.

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9(13+3i)3\left(\frac{1}{3} + 3i\right)^3Show solution

Use (a+b)3=a3+3a2b+3ab2+b3(a+b)^3=a^3+3a^2b+3ab^2+b^3 with a=13a=\frac13, b=3ib=3i.

(13+3i)3\left(\frac13+3i\right)^3

=(13)3+3(13)2(3i)+3(13)(3i)2+(3i)3=\left(\frac13\right)^3+3\left(\frac13\right)^2(3i)+3\left(\frac13\right)(3i)^2+(3i)^3

=127+3⋅19⋅3i+3⋅13⋅9i2+27i3=\frac1{27}+3\cdot\frac1{9}\cdot3i+3\cdot\frac13\cdot9i^2+27i^3

=127+i−9−27i=\frac1{27}+i-9-27i

=−24227−26i=-\frac{242}{27}-26i.

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10(−2−13i)3\left(-2 - \frac{1}{3} i\right)^3Show solution

Let a=−2a=-2 and b=−13b=-\frac13.

Use (a+b)3=a3+3a2b+3ab2+b3(a+b)^3=a^3+3a^2b+3ab^2+b^3:

(−2−13i)3(-2-\frac13 i)^3

=(−2)3+3(−2)2(−13i)+3(−2)(−13i)2+(−13i)3=(-2)^3+3(-2)^2\left(-\frac13 i\right)+3(-2)\left(-\frac13 i\right)^2+\left(-\frac13 i\right)^3

=−8−4i+23i+127i=-8-4i+\frac{2}{3}i+\frac{1}{27}i

=−8−8927i=-8-\frac{89}{27}i.

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114 - 3iShow solution

For z=4−3iz=4-3i, the multiplicative inverse is

z−1=zˉ∣z∣2z^{-1}=\dfrac{\bar z}{|z|^2}.

Here zˉ=4+3i\bar z=4+3i and ∣z∣2=42+(−3)2=16+9=25|z|^2=4^2+(-3)^2=16+9=25.

So,

z−1=4+3i25=45+325iz^{-1}=\dfrac{4+3i}{25}=\dfrac45+\dfrac{3}{25}i.

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125+3i\sqrt{5} + 3iShow solution

For z=5+3iz=\sqrt5+3i, the multiplicative inverse is

z−1=zˉ∣z∣2z^{-1}=\dfrac{\bar z}{|z|^2}.

Here zˉ=5−3i\bar z=\sqrt5-3i and

∣z∣2=(5)2+32=5+9=14|z|^2=(\sqrt5)^2+3^2=5+9=14.

So,

z−1=5−3i14=514−314iz^{-1}=\dfrac{\sqrt5-3i}{14}=\dfrac{\sqrt5}{14}-\dfrac{3}{14}i.

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13-iShow solution

For z=−iz=-i, its multiplicative inverse is

z−1=1−i=iz^{-1}=\frac{1}{-i}=i.

(Equivalent to multiplying numerator and denominator by ii.)

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14Express the following expression in the form of a+ib a + ib :Show solution

First simplify the numerator and denominator:

(3+i5)(3−i5)=32−(i5)2=9−(−5)=14(3+i\sqrt5)(3-i\sqrt5)=3^2-(i\sqrt5)^2=9-(-5)=14.

For the denominator,

(3+2 i)−(3−i2)=3−3+2i+2i=22i(\sqrt3+\sqrt2\,i)-(\sqrt3-i\sqrt2)=\sqrt3-\sqrt3+\sqrt2 i+\sqrt2 i=2\sqrt2 i.

So the expression is

1422i=72i\dfrac{14}{2\sqrt2 i}=\dfrac{7}{\sqrt2 i}.

Rationalising,

72i⋅−2i−2i=−72i2i2=722i\dfrac{7}{\sqrt2 i}\cdot\dfrac{-\sqrt2 i}{-\sqrt2 i}=\dfrac{-7\sqrt2 i}{2i^2}=\dfrac{7\sqrt2}{2}i.

So the value is purely imaginary.

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Miscellaneous Exercise on Chapter 4

1Evaluate: [i18+(1i)25]3\left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^3.
2For any two complex numbers z1z_1 and z2z_2, prove that Re(z1z2)=Rez1Rez2−Imz1Imz2\text{Re}(z_1 z_2) = \text{Re} z_1 \text{Re} z_2 - \text{Im} z_1 \text{Im} z_2.
3Reduce (11−4i−21+i)(3−4i5+i)\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\left(\frac{3-4i}{5+i}\right) to the standard form.
4If x−iy=a−ibc−idx-iy=\sqrt{\frac{a-ib}{c-id}} prove that (x2+y2)2=a2+b2c2+d2\left(x^{2}+y^{2}\right)^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}.
5If z1=2−i,z2=1+iz_{1}=2-i, z_{2}=1+i, find ∣z1+z2+1z1−z2+1∣\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|.
6If a+ib=(x+i)22x2+1a+ib=\frac{(x+i)^{2}}{2x^{2}+1}, prove that a2+b2=(x2+1)2(2x2+1)2a^{2}+b^{2}=\frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}.
7Let z1=2−i,z2=−2+iz_{1}=2-i, z_{2}=-2+i. Find
8Find the real numbers xx and yy if (x−iy)(3+5i)(x-iy)(3+5i) is the conjugate of −6−24i-6-24i.
9Find the modulus of 1+i1−i−1−i1+i\frac{1+i}{1-i}-\frac{1-i}{1+i}.
10If (x+iy)3=u+iv(x+iy)^{3}=u+iv, then show that ux+vy=4(x2−y2)\frac{u}{x}+\frac{v}{y}=4(x^{2}-y^{2}).
11If α\alpha and β\beta are different complex numbers with ∣β∣=1|\beta|=1, then find ∣β−α1−α‾β∣\left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|.
12Find the number of non-zero integral solutions of the equation ∣1−i∣x=2x\left|1-i\right|^{x}=2^{x}.
13If (a+ib)(c+id)(e+if)(g+ih)=A+iB(a+ib)(c+id)(e+if)(g+ih)=A+iB, then show that (a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})(g^{2}+h^{2})=A^{2}+B^{2}.
14If (1+i1−i)m=1\left(\frac{1+i}{1-i}\right)^{m}=1, then find the least positive integral value of mm.

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Complex Numbers and Quadratic Equations covers several key topics that are frequently asked in Madhya Pradesh Board Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
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