Question

The multiplicative inverse of $\frac{3+4i}{4-5i}$ is

• A. $\left(\frac{-8}{25},\frac{31}{25}\right)$
• B. $\left(\frac{-8}{25},\frac{-31}{25}\right)$
• C. $\left(\frac{8}{25},\frac{-31}{25}\right)$
• D. $\left(\frac{8}{25},\frac{31}{25}\right)$

Answer 1

Answer: (B)

Let $z=\frac{3+4i}{4-5i}$
we have to calculate $z^{-1}i.e.,\frac{1}{z}$
$\therefore z^{-1}=\frac{1}{z}=\frac{4-5i}{3+4i}\times \frac{3-4i}{3-4i}$
$=\frac{12-15i-16i+20i^2}{9-16i^2}=\frac{12-31i-20}{9-16\left(-1\right)}$
$=\frac{-8-31i}{9+16}=\frac{-8}{25}-\frac{31}{25}i$
$\therefore z^{-1}=\left(\frac{-8}{25},\frac{-31}{25}\right)$