Question

Find the multiplicative inverse of the following complex numbers :

$\left(i\right)1-i\left(ii\right)(1+i\sqrt{3}{)}^2(iii)4-3i(iv)\sqrt{5}+3i$

Answer 1

(i) 1 - i

If z = x + iy is a complex number, then the multiplicative inverse of z, denoted by $z^{-1}or\frac{1}{z}$ is defined as $z^{-1}=\frac{1}{z}$

$=\frac{1}{x+iy}=\frac{1}{x+iy}\times \frac{x-iy}{x-iy}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}iGiven:z=1-i\therefore z^{-1}=\frac{1}{1^2+1^2}-\frac{(-1)}{1^2+1^2}\times i\frac{1}{2}+\frac{1}{2}i$

$\left(ii\right)(1+i\sqrt{3}{)}^{2}Letz=(1+i\sqrt{3}{)}^2=1^2+(i\sqrt{3}{)}^2+2\times 1\times i\sqrt{3}=1-3+2\sqrt{3}i=-2+2\sqrt{3}i\therefore z^{-1}=\frac{-2}{(-2{)}^2+(2\sqrt{3}{)}^2}-\frac{2\sqrt{3}i}{(-2{)}^2+(2\sqrt{3}{)}^2}=\frac{-2}{4+12}-\frac{2\sqrt{3}i}{4+12}=\frac{-2}{16}-\frac{2\sqrt{3}i}{16}=\frac{-1}{8}-\frac{\sqrt{3}i}{8}$

$\left(iii\right)4-3iLetz=4-3iThenz-1=\frac{4}{4^2+(-3{)}^2}-\frac{(-3)}{4^2+(-3{)}^2}=\frac{4}{16+9}+\frac{3}{16+9}i=\frac{4}{25}+\frac{3}{25}i$

$\left(iv\right)\sqrt{5}+3iLetz=\sqrt{5}+3iThen{z}^{-1}=\frac{\sqrt{5}}{(\sqrt{5}{)}^2+(3{)}^2}-\frac{3}{(\sqrt{5}{)}^2+(3{)}^2}i=\frac{\sqrt{5}}{5+9}-\frac{3}{5+9}i=\frac{\sqrt{5}}{14}-\frac{3}{14}i$