Solve {l}{z=7+i}{a={(z^2+1z^2)}^2}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { z = 7 + i }\\ { a = {(z ^ {2} + 1 z ^ {2})} ^ {2} }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for z, a, b

$$b=8432+5376i$$

Solution Steps

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

$$a=\left(\left(7+i\right)^{2}+1\left(7+i\right)^{2}\right)^{2}$$

Calculate 7+i to the power of 2 and get 48+14i.

$$a=\left(48+14i+1\left(7+i\right)^{2}\right)^{2}$$

Calculate 7+i to the power of 2 and get 48+14i.

$$a=\left(48+14i+1\left(48+14i\right)\right)^{2}$$

Multiply 1 and 48+14i to get 48+14i.

$$a=\left(48+14i+\left(48+14i\right)\right)^{2}$$

Add 48+14i and 48+14i to get 96+28i.

$$a=\left(96+28i\right)^{2}$$

Calculate 96+28i to the power of 2 and get 8432+5376i.

$$a=8432+5376i$$

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

$$b=8432+5376i$$

The system is now solved.

$$z=7+i a=8432+5376i b=8432+5376i$$