Class 11 Maths Ncert Solutions Chapter 5 Ex 5.1 Complex Numbers & Quadratic Equations PDF

# Class 11 Maths Ncert Solutions Ex 5.1

## Class 11 Maths Ncert Solutions Chapter 5 Ex 5.1

Q.1: Express the following complex number in x + iy form; (4i)(74i)$(4i)\;(-\frac{7}{4}i)$

Sol:

(4i)(74i)$\\(4i)\;(-\frac{7}{4}i)$  = -4 × 74$\frac{7}{4}$ × i × i = -7 i2 = -7(-1)   [ Since, i2 = -1 ]

= 7

Q.2: Express the following complex number in x + iy form; i19+i9$i^{19} + i^{9}$

Sol:

i19+i9=i4×2+1+i4×4+3$i^{19} + i^{9}\;=i^{4\times2 + 1} + i^{4\times4 + 3}\\$ = (i4)2i+(i4)4i3$\\(i^{4})^{2}\cdot i + (i^{4})^{4}\cdot i^{3}\\$

=1×i+1×(i)$\\ 1\times i + 1\times (-i)$   [ Since,  i4 = 1, i3 = -1 ]

= i + (-i)

= 0

Q.3: Express the following complex number in x + iy form; i39$i^{-39}$

Sol:

i39=i4×93$i^{-39} = i^{-4\times 9 – 3}\\$ =(i4)9i3$\\= (i^{4})^{-9}\cdot i^{-3}$

=(1)9i3$\\= (1)^{-9}\cdot i^{-3}\\$   [ Since, i4 = 1 ]

=1i3=1i$\\=\frac{1}{i^{3}} = \frac{1}{-i}$   [ Since, i3 = -i ]

=1i×ii$\\= \frac{-1}{i}\times\frac{i}{i}\\$

=ii2=i1$\\= \frac{-i}{i^{2}} = \frac{-i}{-1}$ = i   [ Since, i2 = -1 ]

Q.4: Express the following complex number in x + iy form; 5(6 + 6i) + i(6 + 6i)

Sol:

5(6 + 6i) + i(6 + 6i) = 30 + 30i + 6i + 6i2

= 30 + 36i + 6(-1)    [ Since, i2 = -1 ]

= 30 – 6 + 36i

= 24 + 36i

Q.5: Express the following complex number in x + iy form; (2 – 3i) – (-4 + 5i)

Sol:

(2 – 3i) – (-4 + 5i) = 2 – 3i + 4 – 5i

= 6 – 8i

Q.6: Express the following complex number in x + iy form;

(14+i34)(6+i43)$\left ( \frac{1}{4} + i\frac{3}{4} \right )- \left ( 6 + i\frac{4}{3} \right )$

Sol:

(14+i34)(6+i43)=14+i346i43$\\\left ( \frac{1}{4} + i\frac{3}{4} \right )- \left ( 6 + i\frac{4}{3} \right ) = \frac{1}{4} + i\frac{3}{4} – 6 – i\frac{4}{3}\\$

=(146)+i(3443)$\\\left ( \frac{1}{4} – 6 \right )+ i\left ( \frac{3}{4} – \frac{4}{3} \right )\\$

=234+i(712)$\\\frac{-23}{4} + i(\frac{-7}{12})\\$

=234i712$\boldsymbol{ \frac{-23}{4} – i\frac{7}{12}}$

Q.7: Express the following complex number in x + iy form:

[(13+i73)+(4+i13)](43+i)$\left [ \left ( \frac{1}{3} + i\frac{7}{3} \right ) + \left ( 4 + i\frac{1}{3} \right ) \right ]- \left ( -\frac{4}{3} + i \right )$

Sol:

[(13+i73)+(4+i13)](43+i)$\left [ \left ( \frac{1}{3} + i\frac{7}{3} \right ) + \left ( 4 + i\frac{1}{3} \right ) \right ]- \left ( -\frac{4}{3} + i \right )\\$

=13+i73+4+i13+43i$\\\frac{1}{3} + i\frac{7}{3} + 4 + i\frac{1}{3} + \frac{4}{3} – i\\$

=(13+43+4)+i(73+131)$\\(\frac{1}{3} + \frac{4}{3} + 4) + i(\frac{7}{3} + \frac{1}{3} – 1)\\$

=173+i53$\\\boldsymbol{ \frac{17}{3} + i\frac{5}{3}}$

Q.8: Express the following complex number in x + iy form: (1 – i)4

Sol:

(1 – i)4 = [(1i)2]2$[(1 – i)^{2}]^{2}\\$

=[12i+i2]2$[1 – 2i + i^{2}]^{2}$=[112i]2$[1 – 1 – 2i]^{2}$=(2i)2=(2i)×(2i)=4i2$(-2i)^{2}= (-2i)\times (-2i)= 4i^{2}$          [ Since i2 = -1 ]

= -4

Q.9: Express the following complex number in ‘x + iy’ form; (13+3i)3$(\frac{1}{3} + 3i)^{3}$

Sol:

=(13+3i)3:$(\frac{1}{3} + 3i)^{3}:$

=(13)3+(3i)3+3(13)(3i)(13+3i)$(\frac{1}{3})^{3} + (3i)^{3} + 3(\frac{1}{3})(3i)(\frac{1}{3} +3i)$

=127+27i3+3i(13+3i)$\\\frac{1}{27} + 27i^{3} + 3i(\frac{1}{3} + 3i)$

=127+27i3+i+9i2$\\\frac{1}{27} + 27i^{3} + i + 9i^{2}$

=12727i+i9$\\\frac{1}{27} – 27i + i – 9$      [Since, i3 = -i and i2 = -1]

=(1279)+i(27i+1)$\\(\frac{1}{27} – 9) + i(-27i + 1)\\$ = 2422726i$\boldsymbol{\frac{-242}{27} -26i}$

Q.10: Express the following complex number in ‘x + iy’ form: (2i13)3$(-2 – i\frac{1}{3})^{3}$

SoL:

(2i13)3=(1)3(2+i13)3$(-2 – i\frac{1}{3})^{3}\;=\;(-1)^{3}\;(2 + i\frac{1}{3})^{3}$ = [23+(i3)3+3(2)(i3)(2+i3)]$-[2^{3} + (\frac{i}{3})^{3} +3(2)\;(\frac{i}{3})\;(2 + \frac{i}{3})]$ = [8+i327+2i(2+i3)]$-[8 + \frac{i^{3}}{27} + 2i(2 + \frac{i}{3})]$

=[8+i327+4i+2i23)]$\\-[8 + \frac{i^{3}}{27} + 4i + \frac{2i^{2}}{3})]$ = [8i27+4i23]$-[8 – \frac{i}{27} + 4i – \frac{2}{3}]\\$      [ Since, i3 = -i and i2 = -1 ]

=[223+i10727]$[\frac{22}{3} + i\frac{107}{27}]\\$=221i10727$\boldsymbol{ -\frac{22}{1} – i\frac{107}{27}}$

Q.11: Find the multiplicative inverse of the given complex number, 7 – 6i.

Sol:

Assuming, z = 7 – 6i

Now, z¯¯¯=7+6i$\overline{z} = 7 + 6i$

And, |z|2 = (7)2 + (-6)2 = 49 + 36 = 85

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=7+6i85=785+685i$\\\boldsymbol{z^{-1} = \frac{\overline{z}}{|z|^{2}} \;= \frac{7 + 6i}{85} \;= \frac{7}{85} + \frac{6}{85}i}$

Q.12: Find the multiplicative inverse of the given complex number, 7+4i$\sqrt{7} + 4i$

Sol:

Assuming, z = 7+4i$\sqrt{7} + 4i$

Now, z¯¯¯=7+4i$\overline{z} = \sqrt{7} + 4i$

And, |z|2 = (7)2+(4)2=23$(\sqrt{7})^{2} + (4)^{2} = 23$

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=74i23=723423i$\\\boldsymbol{z^{-1} = \frac{\overline{z}}{|z|^{2}} \;= \frac{\sqrt{7} – 4i}{23} \;= \frac{\sqrt{7}}{23} – \frac{4}{23}i}$

Q.13: Find the multiplicative inverse of the given complex number, – i

Sol:

Assuming, z = – i

Now, z¯¯¯=i$\overline{z} = i$

And, |z|2 = (-i)2 = 1

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=i1$\\z^{-1} = \frac{\overline{z}}{|z|^{2}} = \frac{i}{1}$ = i

Q.14: Express the following complex number in ‘x + iy’ form: (3+i5)(3i5)(3+i2)(3i2)$\frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + i\sqrt{2}) – (\sqrt{3} – i\sqrt{2})}$

Sol:

(3+i5)(3i5)(3+i2)(3i2)=(3)2(i5)23+i23+i2$\boldsymbol{\frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + i\sqrt{2}) – (\sqrt{3} – i\sqrt{2})}}= \frac{(3)^{2} – (i\sqrt{5})^{2}}{\sqrt{3} + i\sqrt{2} – \sqrt{3} + i\sqrt{2}\\}$

Now, by using (a + b) (a – b) = (a – b)2

=95i222i=95×(1)22i=9+522i×ii=14i22i2$= \frac{9 – 5i^{2}}{2\sqrt{2}\;i}= \frac{9 – 5\times (-1)}{2\sqrt{2}\;i}= \frac{9 + 5}{2\sqrt{2}\;i}\times\frac{i}{i}= \frac{14i}{2\sqrt{2}\;i^{2}}\\$

=14i22=7i2×22=72i2$\\\boldsymbol{= \frac{14i}{-2\sqrt{2}}= \frac{-7i}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{-7\sqrt{2}\;i}{2}}$