NCERT Solutions Exercise-4.1 Complex Numbers and Quadratic Equations

NCERT Exercise-4.1 Solutions for Complex Numbers and Quadratic Equations provide detailed step-by-step solutions to important questions based on complex numbers, imaginary numbers, algebraic operations, conjugates, and modulus of complex numbers. These solutions help students strengthen conceptual understanding, improve problem-solving techniques, and prepare effectively for school examinations as well as competitive exams. This exercise is an important part of Class 11 Mathematics and is highly useful for CBSE, JEE Main, JEE Advanced, NDA, and other competitive examination preparation.

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NCERT Exercise 4.1 : Express each of the complex number given in the Questions 1 to 10 in the form a + ib.

Question.1 : Express the complex number in the form a + ib : $(5i)\left(\frac{-3i}{5}\right)$

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,
$(5i)\left(\frac{-3i}{5}\right)=3+i0$

Master related concepts such as SETS NCERT Solutions Exercise 1.4 for Class 11 Maths

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Question 2: Express the complex number in the form a + ib : $i^9+i^{19}$

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,

$i^9+i^{19}=0+i0$

Students should also study SETS NCERT Solutions Miscellaneous Exercise for Class 11 Maths : Maths Anand Classes

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Question 3: Express the complex number in the form a + ib : $i^{-39}$

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,

$i^{-39}=0+i1$

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Question 4: Express the complex number in the form a + ib : $3(7+i7)+i(7+i7)$

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,

$3(7+i7)+i(7+i7)=14+i28$

Build strong concepts by studying Sets Exercise 1.2 NCERT Solutions Class 11 Maths : Maths Anand Classes

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Question 5: Express the complex number in the form a + ib : $(1-i)-(-1+i6)$

Solution:

$(1-i)-(-1+i6)=1-i+1-i6=2-i7$

$(1-i)-(-1+i6)=2-i7$

Important exam-related topics include NCERT Solutions for SETS Exercise 1.3 Class 11 Maths : Maths Anand Classes

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Question 6: Express the complex number in the form a + ib :$(\frac{1}{5}+i\frac{2}{5})-(4+i\frac{5}{2})$

Solution:

$(\frac{1}{5}+i\frac{2}{5})-(4+i\frac{5}{2})=\frac{1}{5}+i\frac{2}{5}-4-i\frac{5}{2}$

$=\frac{1}{5}-4+i\frac{2}{5}-i\frac{5}{2}$

$=-\frac{19}{5}-i\frac{21}{10}$

Hence,

$(\frac{1}{5}+i\frac{2}{5})-(4+i\frac{5}{2})=-\frac{19}{5}-i\frac{21}{10}$

Read More about NCERT Solutions Complex Numbers and Quadratic Equations Miscellaneous Exercise

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Question 7: Express the complex number in the form a + ib :$$[(\frac{1}{3}+i\frac{7}{3})+(4+i\frac{1}{3})]-(-\frac{4}{3}+i)$$

Solution:

$[\frac{1}{3}+i\frac{7}{3}+4+i\frac{1}{3}]+\frac{4}{3}-i$

$=\frac{13}{3}+i\frac{8}{3}+\frac{4}{3}-i$

$=\frac{17}{3}+i\frac{5}{3}$

Hence,

$[(\frac{1}{3}+i\frac{7}{3})+(4+i\frac{1}{3})]-(-\frac{4}{3}+i)=\frac{17}{3}+i\frac{5}{3}$

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Question 8: Express the complex number in the form a + ib : $(1-i{)}^4$

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,
$(1-i{)}^4=-4+i0$

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Question 9: Express the complex number in the form a + ib : ${(\frac{1}{3}+3i)}^3$

Solution:

Using expansion,

${(\frac{1}{3}+3i)}^3={\left(\frac{1}{3}\right)}^3+(3i{)}^3+3{\left(\frac{1}{3}\right)}^2(3i)+3(3i{)}^2\left(\frac{1}{3}\right)$

$=\frac{1}{27}+27i^3+3\cdot \frac{1}{9}\cdot 3i+3\cdot 9i^2\cdot \frac{1}{3}$

$=\frac{1}{27}+27i^3+i+9i^2$

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,

${(\frac{1}{3}+3i)}^3=-\frac{242}{27}-i26$

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Question 10: Express the complex number in the form a + ib : ${(-2-i\frac{1}{3})}^3$

Solution:

Using expansion,

${(-2-i\frac{1}{3})}^3=(-2{)}^3+{(-i\frac{1}{3})}^3+3(-2){(-i\frac{1}{3})}^2+3(-i\frac{1}{3})(-2{)}^2$

$=-8-i^3\frac{1}{27}-6\left(\frac{i^2}{9}\right)-i\cdot 4$

$=-8-i^3\frac{1}{27}-\frac{2i^2}{3}-4i$

Since $i^2=-1$ and $i^3=-i$,

$=-8+i\frac{1}{27}+\frac{2}{3}-4i$

$=-\frac{22}{3}+i(\frac{1}{27}-4)$

$=-\frac{22}{3}-i\frac{107}{27}$

Hence,

${(-2-i\frac{1}{3})}^3=-\frac{22}{3}-i\frac{107}{27}$

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Find the multiplicative inverse of each of the complex numbers given in the Questions 11 to 13.

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Question11 : Find the multiplicative inverse of $4-3i$

Solution

Let $4-3i=z$ (so $\overline{z}=4+3i$).

$|z{|}^2=4^2+(-3{)}^2=16+9=25$

Multiplicative inverse of $z$ is $z^{-1}$.

$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{4+3i}{25}=\frac{4}{25}+i\frac{3}{25}$

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Question 12. Find the multiplicative inverse of $\sqrt{5}+3i$

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 $z^{-1}$.

$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{\sqrt{5}-3i}{14}=\frac{\sqrt{5}}{14}-i\frac{3}{14}$

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Question13. Find the multiplicative inverse of $-i$

Solution

Let $-i=z$ (so $\overline{z}=i$).

$|z{|}^2=(-1{)}^2=1$

Multiplicative inverse of $z$ is $z^{-1}$.

$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{i}{1}=i$

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Question 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

$0-i\frac{7\sqrt{2}}{2}\text{or}-\frac{7\sqrt{2}}{2}i.$

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