Exercise 5.1 complex no. ncert math solution class 11

Exercise 5.1 complex no. ncert math solution class 11

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Exercise 5.1(Complex Number)

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.(Exercise 5.1 complex no. ncert math solution class 11)

Exercise 5.1 complex no. ncert math solution class 11

Question 1: (5 i)\left(-\frac{3}{5} i\right)

Solution: (5 i)\left(-\frac{3}{5} i\right)

=-5 i \times \frac{3}{5} \times i

=-3 i^{2} \quad {\left[\because i^{2}=-1\right]}

=-3(-1)

=3

=3+i 0

Question 2: i^{9}+i^{19}

Solution: i^{9}+i^{19} =i^{4 \times 2+1}+i^{4 \times 4+3}

=\left(i^{4}\right)^{2} \times i+\left(i^{4}\right)^{4} \times i^{3}

=1 \times i+1 \times(-i) \quad\left[\because i^{4}=1, i^{3}=-i\right]

=i+(-i)

=0

=0+i 0

Question 3: i^{-39}

Solution: i^{-39} =i^{4 \times(-9)-3}

=\left(i^{4}\right)^{-9} \times i^{-3}

=(1)^{-9} \times i^{-3} {\left[\because i^{4}=1\right]}

=\frac{1}{i^{3}}

=\frac{1}{-i} {\left[\because i^{3}=-i\right]}

=-\frac{1}{i} \times \frac{i}{i}

=-\frac{i}{i^{2}}

=\frac{-i}{-1} \quad {\left[\because i^{2}=-1\right]}

=i

=0+i

Question 4: 3(7+i 7)+i(7+i 7)

Solution: 3(7+i 7)+i(7+i 7)

=21+21 i+7 i+7 i^{2}

=21+28 i+7 \times(-1) \quad\left[\because i^{2}=-1\right]

=14+28 i

Question 5: (1-i)-(-1+i 6)

Solution: (1-i)-(-1+i 6) =1-i+1-6 i

=2-i 7

Question 6: \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}+\frac{2}{5} i-4-\frac{5}{2} i

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

=\left(-\frac{19}{5}\right)+i\left(-\frac{21}{10}\right)

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

Question 7: \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}+\frac{7}{3} i+4+\frac{1}{3} i+\frac{4}{3}-i

=\left(\frac{1}{3}+4+\frac{4}{3}\right)+i\left(\frac{7}{3}+\frac{1}{3}-1\right)

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

Question 8: (1-i)^{4}

Solution: (1-i)^{4}

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

=\left[1^{2}+i^{2}-2 i\right]^{2}

=[1-1-2 i]^{2} \quad\left(\because i^{2}=-1\right)

=[2 i]^{2}

=4 i^{2} \quad\left(\because i^{2}=-1\right)

=-4

Question 9: \left(\frac{1}{3}+3 i\right)^{3}

Solution: \left(\frac{1}{3}+3 i\right)^{3} =\left(\frac{1}{3}\right)^{3}+(3 i)^{3}+3\left(\frac{1}{3}\right)(3 i)\left(\frac{1}{3}+3 i\right)

=\frac{1}{27}+27 i^{3}+3 i\left(\frac{1}{3}+3 i\right)

=\frac{1}{27}+27(-i)+i+9 i^{2} \quad\left(\because i^{3}=-i\right)

=\frac{1}{27}-27 i+i-9 \quad\left(\because i^{2}=-1\right)

=\left(\frac{1}{27}-9\right)-26 i

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

Question 10: \left(-2-\frac{1}{3} i\right)^{3}

Solution: \left(-2-\frac{1}{3} i\right)^{3} =(-1)^{3}\left(2+\frac{1}{3} i\right)^{3}

=-\left[2^{3}+\left(\frac{i}{3}\right)^{3}+3(2)\left(\frac{i}{3}\right)\left(2+\frac{i}{3}\right)\right]

=-\left[8+\frac{i^{3}}{27}+2 i\left(2+\frac{i}{3}\right)\right] \quad\left[\because i^{3}=-i\right]

=-\left[8-\frac{i}{27}+4 i+\frac{2}{3} i^{2}\right] \quad\left[\because i^{2}=-1\right]

=-\left[8-\frac{i}{27}+4 i-\frac{2}{3}\right]

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

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

Question 11: 4-3 i

Solution: Let z=4-3 i

Therefore, the multiplicative inverse of 4-3 i is = \dfrac{1}{4-3i}

=\dfrac{1}{4-3i}\times \dfrac{4+3i}{4+3i}

=\dfrac{4+3i}{(4)^2-(3i)^2}

=\dfrac{4+3i}{16-9i^2}

=\dfrac{4+3i}{16+9}

=\dfrac{4}{25}+\dfrac{3}{25}i

Question 12: \sqrt{5}+3 i

Solution: Let z=\sqrt{5}+3 i

Then, \bar{z}=\sqrt{5}-3 i and

Therefore, the multiplicative inverse of \sqrt{5}+3 i is =\dfrac{1}{\sqrt{5}+3i}

=\dfrac{1}{\sqrt{5}+3i}\times \dfrac{\sqrt{5}-3i}{\sqrt{5}-3i}

=\dfrac{\sqrt{5}-3i}{(\sqrt{5})^2-(3i)^2}

=\dfrac{\sqrt{5}+3i}{5-9i^2}

=\dfrac{\sqrt{5}-3i}{5+9}\quad[i^2=-1]

=\dfrac{\sqrt{5}-3i}{14}

=\dfrac{\sqrt{5}}{14}-\dfrac{3}{14}i

Question 13: -i

Solution: Let z=-i

Therefore, the multiplicative inverse of -i is =\dfrac{1}{-i}

=-\dfrac{1}{i}\times \dfrac{i}{i}

= -\dfrac{i}{-1}

=i

=0+i

Question 14: Express the following expression in the form of a+i b :

\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{(3)^{2}-(i \sqrt{5})^{2}}{\sqrt{3}+\sqrt{2} i-\sqrt{3}+\sqrt{2} i} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]

=\frac{9-5 i^{2}}{2 \sqrt{2} i}

=\frac{9-5(-1)}{2 \sqrt{2} i} \quad\left[\because i^{2}=-1\right]

=\frac{9+5}{2 \sqrt{2} i}

=\frac{14}{2 \sqrt{2} i} \times \frac{i}{i}

=\frac{7 i}{\sqrt{2} i^{2}}

=\frac{7 i}{\sqrt{2}(-1)}\quad {\left[\because i^{2}=-1\right]}

=\frac{-7 i}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}

=\frac{-7 \sqrt{2} i}{2}

=0-\frac{7 \sqrt{2}}{2}i

Exercise 4.1 ncert math solution class 11

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