Class 11 RD Sharma Solutions – Chapter 13 Complex Numbers – Exercise 13.3

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Question 1. Find the square root of the following complex numbers.

(i) – 5 + 12i

(ii) -7 – 24i

(iii) 1 – i

(iv) – 8 – 6i

(v) 8 – 15i

(vi) $-11 - 60\sqrt-1$

(vii) $1 + 4\sqrt3$

(viii) 4i

(ix) -i

Solution:

If b > 0, $\sqrt{a+ib}=\pm\left[\left(\frac{a+\sqrt{a^2+b^2}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-a+\sqrt{a^2+b^2}}{2}\right)^{\frac{1}{2}}\right]$

If b < 0, $\sqrt{a+ib}=\pm\left[\left(\frac{a+\sqrt{a^2+b^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-a+\sqrt{a^2+b^2}}{2}\right)^{\frac{1}{2}}\right]$

(i) – 5 + 12i

Given:
– 5 + 12i
We know, Z = a + ib
So, $\sqrt{a + ib} = \sqrt{-5+12i}$
Here, b > 0
Let us simplify now,
$\sqrt{-5+12i}=\pm\left[\left(\frac{-5+\sqrt{(-5)^2+12^2}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{(-5)^2+12^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b>0]\\ =\pm\left[\left(\frac{-5+\sqrt{25+144}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{25+144}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-5+\sqrt{169}}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+\sqrt{169}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-5+13}{2}\right)^{\frac{1}{2}}+i\left(\frac{5+13}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8}{2}\right)^{\frac{1}{2}}+i\left(\frac{18}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[4^{\frac{1}{2}}+i9^{\frac{1}{2}}\right]\\ =\pm[2+3i]$
∴ Square root of (– 5 + 12i) is ±[2 + 3i]

(ii) -7 – 24i

Given:
-7 – 24i
We know, Z = -7 – 24i
So, $\sqrt{a + ib} = \sqrt{-7 - 24i}$
Here, b < 0
Let us simplify now,
$\sqrt{-7-24i}=\pm\left[\left(\frac{-7+\sqrt{(-7)^2+(-24)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{(-7)^2+(-24)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{-7+\sqrt{49+576}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{49+576}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-7+\sqrt{625}}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+\sqrt{625}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-7+25}{2}\right)^{\frac{1}{2}}-i\left(\frac{7+25}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{18}{2}\right)^{\frac{1}{2}}-i\left(\frac{32}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[9^{\frac{1}{2}}-i16^{\frac{1}{2}}\right]\\ =\pm[3-4i]$
∴ Square root of (-7 – 24i) is ± [3 – 4i]

(iii) 1 – i

Given:
1 – i
We know, Z = (1 – i)
So, $\sqrt{a + ib} = \sqrt{1 - i}$
Here, b < 0
Let us simplify now,
$\sqrt{1-i}=\pm\left[\left(\frac{1+\sqrt{(1)^2+(-1)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{(1)^2+(-1)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{1+\sqrt{1+1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{1+1}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1+\sqrt{2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-1+\sqrt{2}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\sqrt{\frac{\sqrt2+1}{2}}\right)-i\left(\sqrt{\frac{\sqrt2-1}{2}}\right)\right]$
∴ Square root of (1 – i) is ± $\pm\left[\left(\sqrt{\frac{\sqrt2+1}{2}}\right)-i\left(\sqrt{\frac{\sqrt2-1}{2}}\right)\right]$

(iv) -8 -6i

Given:
-8 -6i
We know, Z = -8 -6i
So, $\sqrt{a + ib}$ = -8 -6i
Here, b < 0
Let us simplify now,
$\sqrt{-8-6i}=\pm\left[\left(\frac{-8+\sqrt{(-8)^2+(-6)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{(-8)^2+(-6)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{-8+\sqrt{64+36}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{64+36}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-8+\sqrt{100}}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+\sqrt{100}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-8+10}{2}\right)^{\frac{1}{2}}-i\left(\frac{8+10}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{2}{2}\right)^{\frac{1}{2}}-i\left(\frac{10}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[1^{\frac{1}{2}}-i9^{\frac{1}{2}}\right]\\ =\pm[1-3i]$
∴ Square root of (-8 -6i) is ± [1 – 3i]

(v) 8 – 15i

Given:
8 – 15i
We know, Z = 8 – 15i
So, $\sqrt{a + ib}$ = 8 – 15i
Here, b < 0
Let us simplify now,
$\sqrt{8-15i}=\pm\left[\left(\frac{8+\sqrt{(8)^2+(-15)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-8+\sqrt{(8)^2+(-15)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{8+\sqrt{64+225}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-8+\sqrt{64+225}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8+\sqrt{289}}{2}\right)^{\frac{1}{2}}-i\left(\frac{-8+\sqrt{289}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8+17}{2}\right)^{\frac{1}{2}}-i\left(\frac{-8+17}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{25}{2}\right)^{\frac{1}{2}}-i\left(\frac{9}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\frac{5}{\sqrt2}-\frac{13}{\sqrt2}\right]\\ =\pm\frac{1}{\sqrt2}(5-3i)$
∴ Square root of (8 – 15i) is ± $\frac{1}{\sqrt2}(5-3i)$

(vi) $-11 - 60\sqrt{-1}$

Given:
$-11 - 60\sqrt{-1}$
We know, Z = $-11 - 60\sqrt{-1}$
So, $\sqrt{a + ib} = -11 - 60\sqrt{-1}$
= -11 – 60i
Here, b < 0
Let us simplify now,
$\sqrt{-11-60i}=\pm\left[\left(\frac{-11+\sqrt{(-11)^2+(-60)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{11+\sqrt{(-11)^2+(60)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{-11+\sqrt{121+3600}}{2}\right)^{\frac{1}{2}}-i\left(\frac{11+\sqrt{121+3600}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-11+\sqrt{3721}}{2}\right)^{\frac{1}{2}}-i\left(\frac{11+\sqrt{3721}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{-11+61}{2}\right)^{\frac{1}{2}}-i\left(\frac{11+61}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{50}{2}\right)^{\frac{1}{2}}-i\left(\frac{72}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[25^{\frac{1}{2}-i36^{\frac{1}{2}}}\right]\\ =\pm(5-6i)$
∴ Square root of ( $-11 - 60\sqrt{-1}$ ) is ± (5 – 6i)

(vii) $1 + 4\sqrt{-3}$

Given:
$1 + 4\sqrt{-3}$
We know, Z = $1 + 4\sqrt{-3}$
So, $\sqrt{a + ib} = 1 + 4\sqrt{-3}$
$= 1 + 4(\sqrt{3}) (\sqrt{-1})\\ = 1 + 4\sqrt{3i}$
Here, b > 0
Let us simplify now,
$\sqrt{1+4\sqrt3i}=\pm\left[\left(\frac{1+\sqrt{(1)^2+(4\sqrt3)^2}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+\sqrt{1^2+(4\sqrt3)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b>0]\\ =\pm\left[\left(\frac{1+\sqrt{1+48}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+\sqrt{1+48}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1+\sqrt{49}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+\sqrt{49}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1+7}{2}\right)^{\frac{1}{2}}+i\left(\frac{-1+7}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{8}{2}\right)^{\frac{1}{2}}+i\left(\frac{6}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[4^{\frac{1}{2}}+i3^{\frac{1}{2}}\right]\\ =\pm[2+\sqrt3i]$
∴ Square root of $(1 + 4\sqrt{-3})$ is ± $(2+\sqrt3i)$

(viii) 4i

Given:
4i
We know, Z = 4i
So, $\sqrt{a + ib}$ = 4i
Here, b > 0
Let us simplify now,
$\sqrt{4i}=\pm\left[\left(\frac{0+\sqrt{(0)^2+(4)^2}}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+\sqrt{0^2+(4)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b>0]\\ =\pm\left[\left(\frac{0+\sqrt{0+16}}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+\sqrt{0+16}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+\sqrt{16}}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+\sqrt{16}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+4}{2}\right)^{\frac{1}{2}}+i\left(\frac{0+4}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{4}{2}\right)^{\frac{1}{2}}+i\left(\frac{4}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[2^{\frac{1}{2}}+i2^{\frac{1}{2}}\right]\\ =\pm[\sqrt2+\sqrt2i]\\ =\pm\sqrt2(1+i)$
∴ Square root of 4i is ± $\sqrt{2 (1 + i)}$

(ix) –i

Given:
-i
We know, Z = -i
So, $\sqrt{(a + ib)}$ = -i
Here, b < 0
Let us simplify now,
$\sqrt{-i}=\pm\left[\left(\frac{0+\sqrt{0^2+(-1)^2}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{0^2+(-1)^2}}{2}\right)^{\frac{1}{2}}\right]\ \ \ [Since\ b<0]\\ =\pm\left[\left(\frac{0+\sqrt{0+1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{0+1}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+\sqrt{1}}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+\sqrt{1}}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{0+1}{2}\right)^{\frac{1}{2}}-i\left(\frac{0+1}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\left(\frac{1}{2}\right)^{\frac{1}{2}}-i\left(\frac{1}{2}\right)^{\frac{1}{2}}\right]\\ =\pm\left[\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right]\\ =\pm\frac{1}{\sqrt2}(1-i)$
∴ Square root of –i is ± $\frac{1}{\sqrt2} (1 - i)$

Last Updated : 21 Feb, 2021

Chapter 1: Sets

Chapter 2: Relations & Functions

Chapter 3: Trigonometric Functions

Chapter 4: Principle of Mathematical Induction

Chapter 5: Complex Numbers and Quadratic Equations

Chapter 6: Linear Inequalities

Chapter 7: Permutations and Combinations

Chapter 8: Binomial Theorem

Chapter 9: Sequences and Series

Chapter 10: Straight Lines

Chapter 11: Conic sections

Chapter 12: Introduction to Three-dimensional Geometry

Chapter 13: Limits and Derivatives

Chapter 14: Mathematical Reasoning

Chapter 15: Statistics

Chapter 16: Probability

NCERT Solutions Chapter 1: Chapter Sets

NCERT Solutions Chapter 2: Relations & Functions

NCERT Solutions Chapter 3: Trigonometric Functions

NCERT Solutions Chapter 4: Principle of Mathematical Induction

NCERT Solutions Chapter 5: Complex Numbers and Quadratic Equations

NCERT Solutions Chapter 6: Linear Inequalities

NCERT Solutions Chapter 7: Permutations and Combinations

NCERT Solutions Chapter 8: Binomial Theorem

NCERT Solutions Chapter 9: Sequences and Series

NCERT Solutions Chapter 10: Straight Lines

NCERT Solutions Chapter 11: Conic sections

NCERT Solutions Chapter 12: Introduction to Three-dimensional Geometry

NCERT Solutions Chapter 13: Limits and Derivatives

NCERT Solutions Chapter 14: Mathematical Reasoning

NCERT Solutions Chapter 15: Statistics

NCERT Solutions Chapter 16: Probability

RD Sharma Solutions Chapter 1: Sets

RD Sharma Solutions Chapter 2: Relations

RD Sharma Solutions Chapter 3: Functions

RD Sharma Solutions Chapter 4: Measurement of Angles

RD Sharma Solutions Chapter 5: Trigonometric Functions

RD Sharma Solutions Chapter 6: Graphs of Trigonometric Functions

RD Sharma Solutions Chapter 7: Trigonometric Ratios of Compound Angles

RD Sharma Solutions Chapter 8: Transformation Formulae

RD Sharma Solutions Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles

RD Sharma Solutions Chapter 10: Sine and Cosine Formulae and their Applications

RD Sharma Solutions Chapter 11: Trigonometric Equations

RD Sharma Solutions Chapter 12: Mathematical Induction

RD Sharma Solutions Chapter 13: Complex Numbers

RD Sharma Solutions Chapter 14: Quadratic Equations

RD Sharma Solutions Chapter 15: Linear Inequations

RD Sharma Solutions Chapter 16: Permutations

RD Sharma Solutions Chapter 17: Combinations

RD Sharma Solutions Chapter 18: Binomial Theorem

RD Sharma Solutions Chapter 19: Arithmetic Progressions

RD Sharma Solutions Chapter 21: Some Special Series

RD Sharma Solutions Chapter 22: Brief Review of Cartesian System of Rectangular

RD Sharma Solutions Chapter 23: The Straight Lines

RD Sharma Solutions Chapter 24: The Circle

RD Sharma Solutions Chapter 25: Parabola

RD Sharma Solutions Chapter 26: Ellipse

RD Sharma Solutions Chapter 27: Hyperbola

RD Sharma Solutions Chapter 28: Introduction to 3D Coordinate Geometry

RD Sharma Solutions Chapter 29: Limits

RD Sharma Solutions Chapter 30: Derivatives

RD Sharma Solutions Chapter 31: Mathematical Reasoning

RD Sharma Solutions Chapter 32: Statistics

RD Sharma Solutions Chapter 33: Probability