Class 11 Maths Exercise 4.1: Complex Numbers
Overview
This page provides comprehensive Class 11 Ch 4: Complex Numbers - Exercise 4.1 Practice. Free NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations Exercise 4-1. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
Complex Numbers: Algebra and Powers of i
About Class 11 Maths Complex Numbers And Quadratic Equations Exercise 4.1
This NCERT Class 11 Maths exercise focuses on Complex Numbers: Algebra and Powers of i from the chapter Complex Numbers And Quadratic Equations. Revise the relevant definitions, standard results, notation, and worked examples before attempting the questions.
Attempt each question independently and write the reasoning in complete steps. Check symbols, substitutions, diagrams, and final conditions carefully, then use the step-by-step solution to identify exactly what needs revision.
Q1
00:00
Express the complex number $5i \cdot (-\frac{3}{5}i)$ in the form $a + ib$.
$5i \cdot (-\frac{3}{5}i) = -3i^2$
Since $i^2 = -1$, we have $-3(-1) = 3$.
In the form $a + ib$, this is $3 + i0$.
$3 + i0$
Q2
00:00
Express the complex number $i^9 + i^{19}$ in the form $a + ib$.
$i^9 = i^{8} \cdot i = (i^4)^2 \cdot i = (1)^2 \cdot i = i$.
$i^{19} = i^{16} \cdot i^3 = (i^4)^4 \cdot (-i) = 1 \cdot (-i) = -i$.
$i^9 + i^{19} = i + (-i) = 0$.
In the form $a + ib$, this is $0 + i0$.
$0 + i0$
Q3
00:00
Express the complex number $i^{-39}$ in the form $a + ib$.
$i^{-39} = \frac{1}{i^{39}} = \frac{1}{i^{36} \cdot i^3} = \frac{1}{1 \cdot (-i)} = \frac{1}{-i}$.
Multiply numerator and denominator by $i$: $\frac{1 \cdot i}{-i \cdot i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$.
In the form $a + ib$, this is $0 + i$.
$0 + i$
Q4
00:00
Express $3(7 + i7) + i(7 + i7)$ in the form $a + ib$.
Expand the terms: $21 + 21i + 7i + 7i^2$.
Substitute $i^2 = -1$: $21 + 28i + 7(-1)$.
$21 - 7 + 28i = 14 + 28i$.
$14 + 28i$
Q5
00:00
Express $(1 - i) - (-1 + i6)$ in the form $a + ib$.
Group real and imaginary parts: $(1 - (-1)) + i(-1 - 6)$.
$1 + 1 + i(-7) = 2 - 7i$.
$2 - 7i$
Q6
00:00
Find the multiplicative inverse of the complex number $4 - 3i$.
Let $z = 4 - 3i$. The multiplicative inverse is $z^{-1} = \frac{1}{4 - 3i}$.
Multiply by the conjugate $\frac{4 + 3i}{4 + 3i}$:
$\frac{4 + 3i}{(4 - 3i)(4 + 3i)} = \frac{4 + 3i}{16 - (3i)^2} = \frac{4 + 3i}{16 - 9i^2}$.
Since $i^2 = -1$, denominator is $16 + 9 = 25$.
$z^{-1} = \frac{4 + 3i}{25} = \frac{4}{25} + i\frac{3}{25}$.
$\frac{4}{25} + i\frac{3}{25}$
Q7
00:00
Find the multiplicative inverse of the complex number $\sqrt{5} + 3i$.
Let $z = \sqrt{5} + 3i$. Then $\bar{z} = \sqrt{5} - 3i$ and $|z|^2 = (\sqrt{5})^2 + 3^2 = 5 + 9 = 14$.
Multiplicative inverse $z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{\sqrt{5} - 3i}{14}$.
$z^{-1} = \frac{\sqrt{5}}{14} - i\frac{3}{14}$.
$\frac{\sqrt{5}}{14} - i\frac{3}{14}$
Q8
00:00
Express $(5 - 3i)^3$ in the form $a + ib$.
Using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:
$(5 - 3i)^3 = 5^3 - 3(5)^2(3i) + 3(5)(3i)^2 - (3i)^3$.
$= 125 - 225i + 15(9i^2) - 27i^3$.
$= 125 - 225i - 135 + 27i = -10 - 198i$.
$-10 - 198i$
Q9
00:00
Express $(\frac{1}{5} + i\frac{2}{5}) - (4 + i\frac{5}{2})$ in the form $a + ib$.
Group real and imaginary parts:
$(\frac{1}{5} - 4) + i(\frac{2}{5} - \frac{5}{2})$
Real part: $\frac{1-20}{5} = -\frac{19}{5}$
Imaginary part: $\frac{4-25}{10} = -\frac{21}{10}$
Result: $-\frac{19}{5} - i\frac{21}{10}$
$-\frac{19}{5} - i\frac{21}{10}$
Q10
00:00
Express $[(\frac{1}{3} + i\frac{7}{3}) + (4 + i\frac{1}{3})] - (-\frac{4}{3} + i)$ in the form $a + ib$.
First sum the terms in brackets:
$(\frac{1}{3} + 4) + i(\frac{7}{3} + \frac{1}{3}) = \frac{13}{3} + i\frac{8}{3}$
Now subtract the last term:
$(\frac{13}{3} + i\frac{8}{3}) - (-\frac{4}{3} + i)$
Group real and imaginary parts:
$(\frac{13}{3} + \frac{4}{3}) + i(\frac{8}{3} - 1)$
$\frac{17}{3} + i(\frac{5}{3})$
$\frac{17}{3} + i\frac{5}{3}$
Q11
00:00
Express $(1 - i)^4$ in the form $a + ib$.
We can write $(1 - i)^4 = [(1 - i)^2]^2$.
Calculate $(1 - i)^2 = 1^2 - 2i + i^2 = 1 - 2i - 1 = -2i$.
Now square the result: $(-2i)^2 = 4i^2 = 4(-1) = -4$.
In $a + ib$ form: $-4 + 0i$.
$-4 + 0i$
Q12
00:00
Express $(\frac{1}{3} + 3i)^3$ in the form $a + ib$.
Using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:
$(\frac{1}{3})^3 + 3(\frac{1}{3})^2(3i) + 3(\frac{1}{3})(3i)^2 + (3i)^3$
$= \frac{1}{27} + 3(\frac{1}{9})(3i) + 1(9i^2) + 27i^3$
$= \frac{1}{27} + i - 9 - 27i$
Group real and imaginary parts:
$(\frac{1}{27} - 9) + i(1 - 27)$
$\frac{1 - 243}{27} - 26i = -\frac{242}{27} - 26i$
$-\frac{242}{27} - 26i$
Q13
00:00
Find the multiplicative inverse of the complex number $-i$.
Let $z = 0 - i$. Then $\bar{z} = 0 + i = i$.
$|z|^2 = 0^2 + (-1)^2 = 1$.
Multiplicative inverse $z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{i}{1} = i$.
In $a + ib$ form: $0 + i$.
$0 + i$
Q14
00:00
Express $\frac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + \sqrt{2}i) - (\sqrt{3} - i\sqrt{2})}$ in the form $a + ib$.
Simplify numerator using $(a+b)(a-b) = a^2 - b^2$:
$3^2 - (i\sqrt{5})^2 = 9 - 5i^2 = 9 - 5(-1) = 9 + 5 = 14$.
Simplify denominator:
$\sqrt{3} + \sqrt{2}i - \sqrt{3} + i\sqrt{2} = 2\sqrt{2}i$.
Expression becomes $\frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i}$.
Rationalize by multiplying numerator and denominator by $i$:
$\frac{7i}{\sqrt{2}i^2} = \frac{7i}{-\sqrt{2}} = -\frac{7}{\sqrt{2}}i$.
Rationalize the radical: $-\frac{7\sqrt{2}}{2}i$.
In $a + ib$ form: $0 - i\frac{7\sqrt{2}}{2}$.
$0 - i\frac{7\sqrt{2}}{2}$