Question

If Z =
$\frac{4 + 3i}{5 - 3i}$
find the value of z^-1​

Answer 1

Answer:

Answer: $\frac{11}{25}-\frac{27}{25}i$

Step-by-step explanation:

Given:

$Z=\frac{4+3i}{5-3i}$

Therefore:

$Z^{-1}=(\frac{5-3i}{4+3i})^{1}=\frac{5-3i}{4+3i}\\ \\Apply\:complex\:arithmetic\:rule}:\quad \frac{a+bi}{c+di}\:=\:\frac{\left(c-di\right)\left(a+bi\right)}{\left(c-di\right)\left(c+di\right)}\:=\:\frac{\left(ac+bd\right)+\left(bc-ad\right)i}{c^2+d^2} \\ \\ a=5,\:b=-3,\:c=4,\:d=3 \\ \\ =\frac{\left(5\cdot \:4+\left(-3\right)\cdot \:3\right)+\left(-3\cdot \:4-5\cdot \:3\right)i}{4^2+3^2}=\frac{11-27i}{25} \\ \\ \\ \text{Rewrite \frac{11-27i}{25} in standard complex form: }=\frac{11}{25}-\frac{27}{25}i$