Complex Numbers
NCERT Class 11 Maths • Exercise 4.1 SolutionsStandard Form: $a + ib$
Q1: Express $(5i)(-\frac{3}{5}i)$
$= 5 \times (-\frac{3}{5}) \times i \times i$
$= -3 \times i^2$
$= -3 \times i^2$
Recall: $i^2 = -1$
$= -3 \times (-1) = 3$
Answer: $3 + 0i$
Q2: Express $i^9 + i^{19}$
$i^9 = (i^4)^2 \times i^1 = 1 \times i = i$
$i^{19} = (i^4)^4 \times i^3 = 1 \times (-i) = -i$
$i^{19} = (i^4)^4 \times i^3 = 1 \times (-i) = -i$
Sum $= i + (-i) = 0$
Answer: $0 + 0i$
Q3: Express $i^{-39}$
$i^{-39} = \frac{1}{i^{39}} = \frac{1}{(i^4)^9 \times i^3}$
$= \frac{1}{1^9 \times (-i)} = \frac{1}{-i}$
$= \frac{1}{1^9 \times (-i)} = \frac{1}{-i}$
Rationalize by multiplying numerator and denominator by $i$
$= \frac{1 \cdot i}{-i \cdot i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = i$
Answer: $0 + i$
Q4: Express $3(7 + i7) + i(7 + i7)$
[Image of adding complex numbers]
Open brackets:
$= 21 + 21i + 7i + 7i^2$
$= 21 + 28i + 7(-1)$
$= 21 + 21i + 7i + 7i^2$
$= 21 + 28i + 7(-1)$
$= 21 – 7 + 28i$
$= 14 + 28i$
$= 14 + 28i$
Answer: $14 + i28$
Q5: Express $(1 – i) – (-1 + i6)$
Group Real and Imaginary parts:
$= (1 – (-1)) + (-i – 6i)$
$= (1 + 1) + i(-1 – 6)$
$= (1 – (-1)) + (-i – 6i)$
$= (1 + 1) + i(-1 – 6)$
Answer: $2 – 7i$
Q6: Express $(\frac{1}{5} + i\frac{2}{5}) – (4 + i\frac{5}{2})$
Real Part: $\frac{1}{5} – 4 = \frac{1-20}{5} = -\frac{19}{5}$
Imaginary Part: $i(\frac{2}{5} – \frac{5}{2}) = i(\frac{4-25}{10}) = i(-\frac{21}{10})$
Imaginary Part: $i(\frac{2}{5} – \frac{5}{2}) = i(\frac{4-25}{10}) = i(-\frac{21}{10})$
Answer: $-\frac{19}{5} – i\frac{21}{10}$
Q7: Simplify $[(\frac{1}{3} + i\frac{7}{3}) + (4 + i\frac{1}{3})] – (-\frac{4}{3} + i)$
Step 1 (Brackets):
Real: $\frac{1}{3} + 4 = \frac{13}{3}$
Imaginary: $i(\frac{7}{3} + \frac{1}{3}) = i(\frac{8}{3})$
Result 1: $\frac{13}{3} + i\frac{8}{3}$
Real: $\frac{1}{3} + 4 = \frac{13}{3}$
Imaginary: $i(\frac{7}{3} + \frac{1}{3}) = i(\frac{8}{3})$
Result 1: $\frac{13}{3} + i\frac{8}{3}$
Step 2 (Subtract):
Real: $\frac{13}{3} – (-\frac{4}{3}) = \frac{17}{3}$
Imaginary: $i(\frac{8}{3} – 1) = i(\frac{5}{3})$
Real: $\frac{13}{3} – (-\frac{4}{3}) = \frac{17}{3}$
Imaginary: $i(\frac{8}{3} – 1) = i(\frac{5}{3})$
Answer: $\frac{17}{3} + i\frac{5}{3}$
Q8: Express $(1 – i)^4$
Strategy: $[(1-i)^2]^2$
$(1-i)^2 = 1^2 + i^2 – 2i = 1 – 1 – 2i = -2i$
$[ -2i ]^2 = 4i^2 = 4(-1) = -4$
Answer: $-4 + 0i$
Q9: Express $(\frac{1}{3} + 3i)^3$
Identity: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$= (\frac{1}{3})^3 + 3(\frac{1}{3})^2(3i) + 3(\frac{1}{3})(3i)^2 + (3i)^3$
$= \frac{1}{27} + 3(\frac{1}{9})(3i) + 9i^2 + 27i^3$
$= \frac{1}{27} + 3(\frac{1}{9})(3i) + 9i^2 + 27i^3$
$= \frac{1}{27} + i – 9 – 27i$
$= (\frac{1}{27} – 9) – 26i = -\frac{242}{27} – 26i$
$= (\frac{1}{27} – 9) – 26i = -\frac{242}{27} – 26i$
Answer: $-\frac{242}{27} – i26$
Q10: Express $(-2 – \frac{1}{3}i)^3$
Factor out -1: $(-1)^3 (2 + \frac{1}{3}i)^3 = -(2 + \frac{i}{3})^3$
Expand $(2 + \frac{i}{3})^3 = 2^3 + 3(2)^2(\frac{i}{3}) + 3(2)(\frac{i}{3})^2 + (\frac{i}{3})^3$
$= 8 + 4i + 6(\frac{-1}{9}) + \frac{-i}{27}$
$= 8 + 4i – \frac{2}{3} – \frac{i}{27}$
$= 8 + 4i + 6(\frac{-1}{9}) + \frac{-i}{27}$
$= 8 + 4i – \frac{2}{3} – \frac{i}{27}$
$= (8 – \frac{2}{3}) + i(4 – \frac{1}{27}) = \frac{22}{3} + i\frac{107}{27}$
Multiply by -1 from start:
Multiply by -1 from start:
Answer: $-\frac{22}{3} – i\frac{107}{27}$
Multiplicative Inverse
Formula: $z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{a – ib}{a^2 + b^2}$
Q11: Multiplicative inverse of $4 – 3i$
Let $z = 4 – 3i$
$\bar{z} = 4 + 3i$
$|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25$
$\bar{z} = 4 + 3i$
$|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25$
Inverse $= \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i$
Answer: $\frac{4}{25} + i\frac{3}{25}$
Q12: Multiplicative inverse of $\sqrt{5} + 3i$
Let $z = \sqrt{5} + 3i \Rightarrow \bar{z} = \sqrt{5} – 3i$
$|z|^2 = (\sqrt{5})^2 + 3^2 = 5 + 9 = 14$
$|z|^2 = (\sqrt{5})^2 + 3^2 = 5 + 9 = 14$
Answer: $\frac{\sqrt{5}}{14} – i\frac{3}{14}$
Q13: Multiplicative inverse of $-i$
Let $z = -i$
$\bar{z} = i$
$|z|^2 = 0^2 + (-1)^2 = 1$
$\bar{z} = i$
$|z|^2 = 0^2 + (-1)^2 = 1$
Inverse $= \frac{i}{1} = i$
Answer: $0 + i$
Fractional Simplification
Q14: Simplify $\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i) – (\sqrt{3}-i\sqrt{2})}$
Numerator: $(3)^2 – (i\sqrt{5})^2 = 9 – (i^2 \cdot 5) = 9 – (-5) = 14$
Denominator: Subtract terms
$= \sqrt{3} + \sqrt{2}i – \sqrt{3} + \sqrt{2}i = 2\sqrt{2}i$
$= \sqrt{3} + \sqrt{2}i – \sqrt{3} + \sqrt{2}i = 2\sqrt{2}i$
$\text{Fraction} = \frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i}$
Multiply by $\frac{i}{i}$ to shift imaginary unit from denominator
$= \frac{7i}{\sqrt{2}i^2} = \frac{7i}{-\sqrt{2}} = -\frac{7\sqrt{2}}{2}i$
Answer: $0 – i\frac{7\sqrt{2}}{2}$