Solve {l}{z={(2+3i)}{(1-i)}}{a=z^-1}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { z = {(2 + 3 i)} {(1 - i)} }\\ { a = z ^ {-1} }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for z, a, b

$$b=\frac{5}{26}-\frac{1}{26}i\approx 0.192307692-0.038461538i$$

Solution Steps

Consider the first equation. Multiply 2+3i and 1-i to get 5+i.

$$z=5+i$$

Consider the second equation. Insert the known values of variables into the equation.

$$a=\left(5+i\right)^{-1}$$

Calculate 5+i to the power of -1 and get \frac{5}{26}-\frac{1}{26}i.

$$a=\frac{5}{26}-\frac{1}{26}i$$

Consider the third equation. Insert the known values of variables into the equation.

$$b=\frac{5}{26}-\frac{1}{26}i$$

The system is now solved.

$$z=5+i a=\frac{5}{26}-\frac{1}{26}i b=\frac{5}{26}-\frac{1}{26}i$$