NCERT Solutions for Class 11 Maths Chapter 4 Exercise 4.1 - Complex Numbers and Quadratic Equations

Question 1: Express the given complex number in the form a+ib: (5i)(-3i/5)?
Answer:

Given $a+ib:(5i)\times (-\frac{3}{5}i)$

$(5i)\times (-\frac{3}{5}i)=-(5\times \frac{3}{5})\left(i^2\right)$

$\begin{array}{rl}& =-3\left(i^2\right)\\ & =3.\end{array}$

Question 2: Express each of the complex number in the form $a+ib$ .

$(5i)(-3i/5)$?

Answer:

We know that $i^4=1$
Now, we will reduce $i^9+i^{19}$ into

$i^9+i^{19}={\left(i^4\right)}^2\cdot i+{\left(i^4\right)}^3\cdot i^3$
$=(1{)}^2.i+(1{)}^3.(-i)$ $(\because i^4=1,i^3=-iand{i}^2=-1)$
$=i-i=0$
Now, in the form of $a+ib$ we can write it as
$o+io$
Therefore, the answer is $o+io$

Question 3: Express each of the complex number in the form a+ib.

$i^{-39}$

Answer:

We know that $i^4=1$
Now, we will reduce $i^{-39}$ into

$i^{-39}$ $=(i^4{)}^{-9}.i^{-3}$
$=(1{)}^{-9}.(-i{)}^{-1}$ $(\because i^4=1,i^3=-i)$
$=\frac{1}{-i}$
$=\frac{1}{-i}\times \frac{i}{i}$
$=\frac{i}{-i^2}$ $(\because i^2=-1)$
$=\frac{i}{-(-1)}$
$=i$
Now, in the form of $a+ib$ we can write it as
$o+i1$
Therefore, the answer is $o+i1$

Question 4: Express each of the complex number in the form a+ib.

$3(7+7i)+i(7+7i)$

Answer:

Given problem is
$3(7+7i)+i(7+7i)$
Now, we will reduce it into

$3(7+7i)+i(7+7i)$ $=21+21i+7i+7i^2$
$=21+21i+7i+7(-1)$ $(\because i^2=-1)$
$=21+21i+7i-7$
$=14+28i$

Therefore, the answer is $14+i28$

Question 5: Express each of the complex number in the form $a+ib$ .

$(1-i)-(-1+6i)$

Answer:

Given problem is
$(1-i)-(-1+6i)$
Now, we will reduce it into

$(1-i)-(-1+6i)$$=1-i+1-6i$
$=2-7i$

Therefore, the answer is $2-7i$

Question 6: Express each of the complex number in the form $a+ib$ .

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

Answer:

Given problem is
$(\frac{1}{5}+i\frac{2}{5})-(4+i\frac{5}{2})$
Now, we will reduce it into

$(\frac{1}{5}+i\frac{2}{5})-(4+i\frac{5}{2})=\frac{1}{5}+i\frac{2}{5}-4-i\frac{5}{2}$
$=\frac{1-20}{5}+i\frac{(4-25)}{10}$
$=-\frac{19}{5}-i\frac{21}{10}$

Therefore, the answer is $-\frac{19}{5}-i\frac{21}{10}$

Question 7: Express each of the complex number in the form $a+ib$ .

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

Answer:

Given problem is
$[(\frac{1}{3}+i\frac{7}{3})+(4+i\frac{1}{3})]-(-\frac{4}{3}+i)$
Now, we will reduce it into

$[(\frac{1}{3}+i\frac{7}{3})+(4+i\frac{1}{3})]-(-\frac{4}{3}+i)=\frac{1}{3}+i\frac{7}{3}+4+i\frac{1}{3}+\frac{4}{3}-i$
$=\frac{1+4+12}{3}+i\frac{(7+1-3)}{3}$
$=\frac{17}{3}+i\frac{5}{3}$

Therefore, the answer is $\frac{17}{3}+i\frac{5}{3}$

Question 8: Express each of the complex number in the form $a+ib$ .

$(1-i{)}^4$

Answer:

The given problem is
$(1-i{)}^4$
Now, we will reduce it into

$(1-i{)}^4=((1-i{)}^2{)}^2$
$=(1^2+i^2-2.1.i{)}^2$ $(using(a-b{)}^2=a^2+b^2-2ab)$
$=(1-1-2i{)}^2$ $(\because i^2=-1)$
$=(-2i{)}^2$
$=4i^2$
$=-4$

Therefore, the answer is $-4+i0$

Question 9: Express each of the complex number in the form $a+ib$ .

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

Answer:

Given problem is
${(\frac{1}{3}+3i)}^3$
Now, we will reduce it into

${(\frac{1}{3}+3i)}^3={\left(\frac{1}{3}\right)}^3+(3i{)}^3+3.{\left(\frac{1}{3}\right)}^2.3i+3.\frac{1}{3}.(3i{)}^2$ $(using(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2)$
$=\frac{1}{27}+27i^3+i+9i^2$
$=\frac{1}{27}+27(-i)+i+9(-1)$ $(\because i^3=-iand{i}^2=-1)$
$=\frac{1}{27}-27i+i-9$
$=\frac{1-243}{27}-26i$
$=-\frac{242}{27}-26i$

Therefore, the answer is

$-\frac{242}{27}-26i$

Question 10: Express each of the complex number in the form $a+ib$ .

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

Answer:

Given problem is
${(-2-\frac{1}{3}i)}^3$
Now, we will reduce it into

${(-2-\frac{1}{3}i)}^3=-((2{)}^3+{\left(\frac{1}{3}i\right)}^3+3.(2{)}^2\frac{1}{3}i+3.{\left(\frac{1}{3}i\right)}^2.2)$ $(using(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2)$
$=-(8+\frac{1}{27}{i}^3+3.4.\frac{1}{3}i+3.\frac{1}{9}{i}^2.2)$
$=-(8+\frac{1}{27}(-i)+4i+\frac{2}{3}(-1))$ $(\because i^3=-iand{i}^2=-1)$
$=-(8-\frac{1}{27}i+4i-\frac{2}{3})$
$=-(\frac{(-1+108)}{27}i+\frac{24-2}{3})$
$=-\frac{22}{3}-i\frac{107}{27}$

Therefore, the answer is $-\frac{22}{3}-i\frac{107}{27}$

Question 11: Find the multiplicative inverse of each of the complex numbers.

$4-3i$

Answer:

Let $z=4-3i$
Then,
$\overline{z}=4+3i$
And
$|z{|}^2=4^2+(-3{)}^2=16+9=25$
Now, the multiplicative inverse is given by
$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{4+3i}{25}=\frac{4}{25}+i\frac{3}{25}$

Therefore, the multiplicative inverse is

$\frac{4}{25}+i\frac{3}{25}$

Question 12: Find the multiplicative inverse of each of the complex numbers.

$\sqrt{5}+3i$

Answer:

Let $z=\sqrt{5}+3i$
Then,
$\overline{z}=\sqrt{5}-3i$
And
$|z{|}^2=(\sqrt{5}{)}^2+(3{)}^2=5+9=14$
Now, the multiplicative inverse is given by
$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{\sqrt{5}-3i}{14}=\frac{\sqrt{5}}{14}-i\frac{3}{14}$

Therefore, the multiplicative inverse is $\frac{\sqrt{5}}{14}-i\frac{3}{14}$

Question 13: Find the multiplicative inverse of each of the complex numbers.

$-i$

Answer:

Let $z=-i$
Then,
$\overline{z}=i$
And
$|z{|}^2=(0{)}^2+(1{)}^2=0+1=1$
Now, the multiplicative inverse is given by
$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{i}{1}=0+i$

Therefore, the multiplicative inverse is $0+i1$

Question 14: Express the following expression in the form of $a+ib:$

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$

Answer:

Given problem is
$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$
Now, we will reduce it into

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}=\frac{3^2-(\sqrt{5}i{)}^2}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$ $(using(a-b)(a+b)=a^2-b^2)$
$=\frac{9-5i^2}{\sqrt{3}+\sqrt{2}i-\sqrt{3}+\sqrt{2}i}$
$=\frac{9-5(-1)}{2\sqrt{2}i}$ $(\because i^2=-1)$
$=\frac{14}{2\sqrt{2}i}\times \frac{\sqrt{2}i}{\sqrt{2}i}$
$=\frac{7\sqrt{2}i}{2i^2}$
$=-\frac{7\sqrt{2}i}{2}$
Therefore, the answer is $0-i\frac{7\sqrt{2}}{2}$