Solve z=20t/3-i-(5-3i)*(2+3i)^2+(1+i)^5

$$z = \frac { 20 t } { 3 - i } - ( 5 - 3 i ) \cdot ( 2 + 3 i ) ^ { 2 } + ( 1 + i ) ^ { 5 }$$

Solve for t

$$t=\left(\frac{3}{20}-\frac{1}{20}i\right)z+\left(\frac{31}{5}+\frac{111}{10}i\right)$$

Steps for Solving Linear Equation

Divide 20t by 3-i to get \left(6+2i\right)t.

$$z=\left(6+2i\right)t-\left(5-3i\right)\left(2+3i\right)^{2}+\left(1+i\right)^{5}$$

Calculate 2+3i to the power of 2 and get -5+12i.

$$z=\left(6+2i\right)t-\left(5-3i\right)\left(-5+12i\right)+\left(1+i\right)^{5}$$

Multiply 5-3i and -5+12i to get 11+75i.

$$z=\left(6+2i\right)t-\left(11+75i\right)+\left(1+i\right)^{5}$$

Calculate 1+i to the power of 5 and get -4-4i.

$$z=\left(6+2i\right)t-\left(11+75i\right)+\left(-4-4i\right)$$

Swap sides so that all variable terms are on the left hand side.

$$\left(6+2i\right)t-\left(11+75i\right)+\left(-4-4i\right)=z$$

Add 4+4i to both sides.

$$\left(6+2i\right)t-\left(11+75i\right)=z+\left(4+4i\right)$$

Add 11+75i to both sides.

$$\left(6+2i\right)t=z+\left(4+4i\right)+\left(11+75i\right)$$

Do the additions in 4+4i+\left(11+75i\right).

$$\left(6+2i\right)t=z+15+79i$$

The equation is in standard form.

$$\left(6+2i\right)t=z+\left(15+79i\right)$$

Divide both sides by 6+2i.

$$\frac{\left(6+2i\right)t}{6+2i}=\frac{z+\left(15+79i\right)}{6+2i}$$

Dividing by 6+2i undoes the multiplication by 6+2i.

$$t=\frac{z+\left(15+79i\right)}{6+2i}$$

Divide z+\left(15+79i\right) by 6+2i.

$$t=\left(\frac{3}{20}-\frac{1}{20}i\right)z+\left(\frac{31}{5}+\frac{111}{10}i\right)$$

Solve for z

$$z=\left(6+2i\right)t+\left(-15-79i\right)$$