11th NCERT/CBSE Sequences and series Exercise 9.1 Questions 14
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## Exercise 5.1 Questions 14

Question (1)

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

Solution

$\left( {5i} \right)\left( { - \frac{3}{5}i} \right)$ $= - 3{i^2}$ $= - 3\left( { - 1} \right) = 3$

Question (2)

Express the given complex number in the form a + ib: i9 + i19

Solution

${i^9} + {i^{19}} = {\left( {{i^2}} \right)^4}i + {\left( {{i^2}} \right)^9}i$ $= {\left( { - 1} \right)^4}i + {\left( { - 1} \right)^9}i$ $= i - i = 0$

Question (3)

Express the given complex number in the form a + ib: i–39

Solution

${i^{ - 39}} = {i^{ - 40}}.i$ $= {\left( {{i^4}} \right)^{ - 10}}i$ $= {\left( 1 \right)^{ - 10}}i$ $= i$

Question (4)

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

Solution

3(7 + i7) + i(7 + i7)
= 21 + 21i + 7i + 7i2
= 21 - 28i - 7 [ i2 = -1]
= 14 + 28i

Question (5)

Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)

Solution

(1 – i) – (–1 + i6)
= 1 -i + 1 - 6i
= 2 - 7i

Question (6)

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

Solution

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

Question (7)

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

Solution

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

Question (8)

Express the given complex number in the form a + ib: (1 – i)4

Solution

${\left( {1 - i} \right)^4} = {\left[ {{{\left( {1 - i} \right)}^2}} \right]^2}$ $= {\left( {1 - 2i + {i^2}} \right)^2}$ $= {\left( {1 + 2i - 1} \right)^2}$ $= {\left( {2i} \right)^2} = 4{i^2}$ $= - 4$

Question (9)

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

Solution

${\left( {\frac{1}{3} + 3i} \right)^3}$ $= {\left( {\frac{1}{3}} \right)^3} + {\left( {3i} \right)^3} + 3\left( {\frac{1}{3}} \right)\left( {3i} \right)\left( {\frac{1}{3} + 3i} \right)$ $= \frac{1}{{27}} + 27{i^3} + 3i\left( {\frac{1}{3} + 3i} \right)$ $= \frac{1}{{27}} + 27{i^2}i + i + 9{i^2}$ $= \frac{1}{{27}} - 27i + i - 9$ $= \frac{{1 - 243}}{{27}} - 26i$ $= - \frac{{242}}{{27}} - 26i$

Question (10)

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

Solution

${\left( { - 2 - \frac{1}{3}i} \right)^{ 3}}$ $= {\left[ { - \left( {2 + \frac{1}{3}i} \right)} \right]^3}$ $= - \left[ {8 + \frac{1}{{27}}{i^3} + 2i\left( {2 + \frac{1}{3}i} \right)} \right]$ $= - \left[ {8 + \frac{1}{{27}}i.{i^2} + 4i + \frac{2}{3}{i^2}} \right]$ $= - \left[ {8 - \frac{2}{3} + 4i - \frac{1}{{27}}i} \right]$ $= - \frac{{22}}{3} - \frac{{107}}{{27}}i$

Question (11)

Find the multiplicative inverse of the complex number 4 – 3i

Solution

4 - 3i
Comparing to standard form a = 4, b = -3
a2 + b2 = 42+(-3)2
= 16 + 9 = 25
The multiplicative inverse of a + bi is $\frac{a}{{{a^2} + {b^2}}} - \frac{b}{{{a^2} + {b^2}}}i$ So multiplicative inverse of 4 - 3i is $= \frac{4}{{25}} + \frac{3}{{25}}i$

Question (12)

Find the multiplicative inverse of the complex number $\sqrt 5 + 3i$

Solution

$\sqrt 5 + 3i$ Comparing to standard form a = √5, b = 3
a2 + b2 = √52+(3)2
= 5 + 9 = 14
The multiplicative inverse of a + bi is $\frac{a}{{{a^2} + {b^2}}} - \frac{b}{{{a^2} + {b^2}}}i$ So multiplicative inverse of$\sqrt 5 + 3i$ is $= \frac{{\sqrt 5 }}{{14}} - \frac{3}{{14}}i$

Question (13)

Find the multiplicative inverse of the complex number - i

Solution

- i
Comparing to standard form a = 0, b = -1
a2 + b2 = 02+(-1)2
= 0 + 1 = 1
The multiplicative inverse of a + bi is $\frac{a}{{{a^2} + {b^2}}} - \frac{b}{{{a^2} + {b^2}}}i$ So multiplicative inverse of -i is i

Question (14)

Express the following expression in the form of a + ib. $\frac{{\left( {3 + i\sqrt 5 } \right)\left( {3 - i\sqrt 5 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 i} \right) - \left( {\sqrt 3 - i\sqrt 2 } \right)}}$

Solution

$\frac{{\left( {3 + i\sqrt 5 } \right)\left( {3 - i\sqrt 5 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 i} \right) - \left( {\sqrt 3 - i\sqrt 2 } \right)}}$ $= \frac{{9 - 5{i^2}}}{{\sqrt 3 + \sqrt 2 i - \sqrt 3 + \sqrt 2 i}}$ $\frac{{9 + 5}}{{2\sqrt 2 i}}$ $= \frac{{14\sqrt 2 i}}{{2\sqrt 2 \sqrt 2 {i^2}}}$ $= \frac{{7\sqrt 2 }}{2}i$