Anand Classes offers NCERT Solutions for Class 11 Mathematics – Complex Numbers & Quadratic Equations (Exercise 4.1) with step-by-step explanations, solved examples, and clear diagrams. These solutions cover all questions from Ex 4.1 (Express in form a + ib, finding inverse, operations, powers of i, etc.), following the latest CBSE / NCERT syllabus. Students can download the entire solutions in free PDF format for offline study and exam revision. Click the print button to download study material and notes.
NCERT Exercise 4.1 : Express each of the complex number given in the Questions 1 to 10 in the form a + ib.
Question.1 : $(5i)\left(\frac{-3i}{5}\right)$
Solution:
$5i \times \left(\frac{-3i}{5}\right) = -3 \times i^2$
Since, $i^2 = -1$. Therefore,
$-3 \times i^2 = -3 \times (-1) = 3$
Hence,
$$(5i)\left(\frac{-3i}{5}\right) = 3 + i0$$
Question 2: $i^9 + i^{19}$
Solution:
$i^9 + i^{19} = i(i^2)^4 + i(i^2)^9$
Since, $i^2 = -1$. Therefore,
$i(i^2)^4 + i(i^2)^9 = i(-1)^4 + i(-1)^9$
$= i – i = 0$
Hence,
$$i^9 + i^{19} = 0 + i0$$
Question 3: $i^{-39}$
Solution:
$i^{-39} = \frac{1}{i^{39}} = \frac{1}{i^3 \cdot (i^4)^9}$
Since, $i^3 = -i$ and $i^4 = 1$. Therefore,
$$\frac{1}{i^3 \cdot (i^4)^9} = \frac{1}{(-i)(1)} = \frac{1}{-i}$$
Multiply and divide by $i$:
$$\frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2}$$
Since, $i^2 = -1$ and $-i^2 = 1$ Therefore,
$$\frac{i}{-i^2} = i$$
Hence,
$$i^{-39} = 0 + i1$$
Question 4: $3(7 + i7) + i(7 + i7)$
Solution:
$3(7 + i7) + i(7 + i7) = 21 + i21 + i7 + i^2 7$
Since, $i^2 = -1$. Therefore,
$$21 + i21 + i7 + i^2 7 = 21 + i28 + (-1)7 = 21 – 7 + i28$$
$$= 14 + i28$$
Hence,
$$3(7 + i7) + i(7 + i7) = 14 + i28$$
Question 5: $(1 – i) – (-1 + i6)$
Solution:
$(1 – i) – (-1 + i6) = 1 – i + 1 – i6 = 2 – i7$
$$(1 – i) – (-1 + i6)= 2 – i7$$
Question 6: $\left(\frac{1}{5} + i\frac{2}{5}\right) – (4 + i\frac{5}{2})$
Solution:
$\left(\frac{1}{5} + i\frac{2}{5}\right) – (4 + i\frac{5}{2}) = \frac{1}{5} + i\frac{2}{5} – 4 – i\frac{5}{2}$
$$= \frac{1}{5} – 4 + i\frac{2}{5} – i\frac{5}{2}$$
$$= -\frac{19}{5} – i\frac{21}{10}$$
Hence,
$$\left(\frac{1}{5} + i\frac{2}{5}\right) – (4 + i\frac{5}{2}) = -\frac{19}{5} – i\frac{21}{10}$$
Question 7: $$\left[\left(\frac{1}{3}+i\frac{7}{3}\right)+(4+i\frac{1}{3})\right]-\left(-\frac{4}{3}+i\right)$$
Solution:
$$\left[\frac{1}{3}+i\frac{7}{3}+4+i\frac{1}{3}\right]+ \frac{4}{3}-i$$
$$= \frac{13}{3} + i\frac{8}{3} + \frac{4}{3} – i$$
$$= \frac{17}{3} + i\frac{5}{3}$$
Hence,
$$\left[\left(\frac{1}{3}+i\frac{7}{3}\right)+(4+i\frac{1}{3})\right]-\left(-\frac{4}{3}+i\right)=\frac{17}{3}+i\frac{5}{3}$$
Question 8: $(1-i)^4$
Solution:
$$(1-i)^4=(1-i)^2(1-i)^2$$
$$(1-i)^2 = 1 + i^2 – 2i$$
Since $i^2=-1$,
$$(1-i)^2 = 1 -1 -2i = -2i$$
So
$$(1-i)^4 = (-2i)^2 = 4i^2 = 4(-1) = -4$$
Hence,
$$(1-i)^4 = -4 + i0$$
Question 9: $\left(\frac{1}{3}+3i\right)^3$
Solution:
Using expansion,
$$\left(\frac{1}{3}+3i\right)^3 = \left(\frac{1}{3}\right)^3 + (3i)^3 + 3\left(\frac{1}{3}\right)^2(3i) + 3(3i)^2\left(\frac{1}{3}\right)$$
$$= \frac{1}{27} + 27i^3 + 3\cdot\frac{1}{9}\cdot 3i + 3\cdot 9 i^2 \cdot \frac{1}{3}$$
$$= \frac{1}{27} + 27i^3 + i + 9i^2$$
Since $i^2=-1$ and $i^3=-i$,
$$= \frac{1}{27} -27i + 9(-1) + i$$
$$= \frac{1}{27} -9 -26i$$
$$= -\frac{242}{27} – 26i$$
Hence,
$$\left(\frac{1}{3}+3i\right)^3 = -\frac{242}{27} – i26$$
Question 10: $\left(-2 – i\frac{1}{3}\right)^3$
Solution:
Using expansion,
$$\left(-2 – i\frac{1}{3}\right)^3 = (-2)^3 + \left(-i\frac{1}{3}\right)^3 + 3(-2)\left(-i\frac{1}{3}\right)^2 + 3\left(-i\frac{1}{3}\right)(-2)^2$$
$$= -8 – i^3\frac{1}{27} – 6\left(\frac{i^2}{9}\right) – i\cdot 4$$
$$= -8 – i^3\frac{1}{27} – \frac{2 i^2}{3} – 4i$$
Since $i^2=-1$ and $i^3=-i$,
$$= -8 + i\frac{1}{27} + \frac{2}{3} – 4i$$
$$= -\frac{22}{3} + i\left(\frac{1}{27} – 4\right)$$
$$= -\frac{22}{3} – i\frac{107}{27}$$
Hence,
$$\left(-2 – i\frac{1}{3}\right)^3 = -\frac{22}{3} – i\frac{107}{27}$$
Find the multiplicative inverse of each of the complex numbers given in the Questions 11 to 13.
Question11 : $4 – 3i$
Solution
Let $4 – 3i = z$ (so $\overline{z} = 4 + 3i$).
$$
|z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25
$$
Multiplicative inverse of $z$ is $z^{-1}$.
$$
z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{4 + 3i}{25} = \frac{4}{25} + i\frac{3}{25}
$$
12. $\sqrt{5} + 3i$
Solution
Let $\sqrt{5} + 3i = z$ (so $\overline{z} = \sqrt{5} – 3i$).
$$
|z|^2 = (\sqrt{5})^2 + 3^2 = 5 + 9 = 14
$$
Multiplicative inverse of $z$ is $z^{-1}$.
$$
z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{\sqrt{5} – 3i}{14} = \frac{\sqrt{5}}{14} – i\frac{3}{14}
$$
13. $-i$
Solution
Let $-i = z$ (so $\overline{z} = i$).
$$
|z|^2 = (-1)^2 = 1
$$
Multiplicative inverse of $z$ is $z^{-1}$.
$$
z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{i}{1} = i
$$
14. Express the following expression in the form $a+ib$:
$$
\frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + \sqrt{2}\,i) – (\sqrt{3} – \sqrt{2}\,i)}
$$
Solution
$$
\frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + \sqrt{2}i) – (\sqrt{3} – \sqrt{2}i)}
= \frac{9 – (i\sqrt{5})^2}{\sqrt{3} + \sqrt{2}i – \sqrt{3} + \sqrt{2}i}
= \frac{9 – (i\sqrt{5})^2}{2\sqrt{2}\,i}
$$
Since $i^2 = -1$, we have $(i\sqrt{5})^2 = i^2\cdot 5 = -5$, so
$$
9 – (i\sqrt{5})^2 = 9 – (-5) = 14
$$
Thus
$$
\frac{9 – (i\sqrt{5})^2}{2\sqrt{2}\,i} = \frac{14}{2\sqrt{2}\,i} = \frac{7}{\sqrt{2}\,i}
$$
Multiply numerator and denominator by $\sqrt{2}\,i$:
$$
\frac{7}{\sqrt{2}\,i}\cdot\frac{\sqrt{2}\,i}{\sqrt{2}\,i}
= \frac{7\sqrt{2}\,i}{(\sqrt{2}\,i)^2}
= \frac{7\sqrt{2}\,i}{2i^2}
$$
Since $i^2 = -1$,
$$
\frac{7\sqrt{2}\,i}{2i^2} = \frac{7\sqrt{2}\,i}{2(-1)} = -\frac{7\sqrt{2}}{2}\,i
$$
So the expression in the form $a+ib$ is
$$
0 – i\frac{7\sqrt{2}}{2}\quad\text{or}\quad -\frac{7\sqrt{2}}{2}\,i.
$$
