Class 11th

11. Let Z = 4 – 3i

Then $\overline{z}$ = 4 + 3i

And, z|² = 4²+ (-3)²= 16 + 9 = 25

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

10.${\left(-2-\frac{1}{3}i\right)}^3$

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

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

= $(-8)\left[8+\frac{i^3}{27}+3.2.\frac{1}{3}i\left(2+\frac{i}{3}\right)\right]$ [since, (a + b)³ = a³ + b³ + 3ab(a + b)]

= (–1) $\left[8-\frac{i}{27}+(2i*2)+\left(2i*\frac{i}{3}\right)\right]$ [since, i³ = i².i = –i and i² = –1]

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

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

= $-\frac{22}{3}$ $–\frac{i107}{27}$

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

= $\frac{1}{3}$ + $i\frac{7}{3}$ + 4 + $i\frac{1}{3}$ + $\frac{4}{3}$ – i

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

= $\frac{1+12+4}{3}$ + $i\frac{7+1-3}{3}$

= $\frac{17}{3}$ + $i\frac{5}{3}$

6. $\left(\frac{1}{5}+i\frac{2}{5}\right)$ – $\left(4+i\frac{5}{2}\right)$

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

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

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

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

= $\frac{-19}{5}$ – i $\frac{21}{10}$

5. (1 – i) – (–1 + i6)

= 1 – I + 1 – i6

= 2 – 7i

4. 3 (7 + i7) + I (7 + i7)

= 21 + 21i + 7i + 7i²

= 21 + 28i + 7 (–1) [since i² = –1]

= 21 + 28i – 7

= 14 + 28i

1. (5i) $\left(\frac{-3}{5}i\right)$

= 5 $\left(-\frac{3}{5}\right)$ *i²          [since i² = –1]

= –3 (–1)

= 3

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

61. Kindly go through the solution

60. Kindly go through the solution