Solve z^2-(1-i)^6z=-20

$$z ^ { 2 } - ( 1 - i ) ^ { 6 } z = - 20$$

Solve for z

$$z=10i$$

$$z=-2i$$

Steps Using the Quadratic Formula

Calculate 1-i to the power of 6 and get 8i.

$$z^{2}-8iz=-20$$

Add 20 to both sides.

$$z^{2}-8iz+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8i for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{8i±\sqrt{\left(-8i\right)^{2}-4\times 20}}{2}$$

Square -8i.

$$z=\frac{8i±\sqrt{-64-4\times 20}}{2}$$

Multiply -4 times 20.

$$z=\frac{8i±\sqrt{-64-80}}{2}$$

Add -64 to -80.

$$z=\frac{8i±\sqrt{-144}}{2}$$

Take the square root of -144.

$$z=\frac{8i±12i}{2}$$

Now solve the equation z=\frac{8i±12i}{2} when ± is plus. Add 8i to 12i.

$$z=\frac{20i}{2}$$

Divide 20i by 2.

$$z=10i$$

Now solve the equation z=\frac{8i±12i}{2} when ± is minus. Subtract 12i from 8i.

$$z=\frac{-4i}{2}$$

Divide -4i by 2.

$$z=-2i$$

The equation is now solved.

$$z=10i z=-2i$$

Steps for Completing the Square

Calculate 1-i to the power of 6 and get 8i.

$$z^{2}-8iz=-20$$

Divide -8i, the coefficient of the x term, by 2 to get -4i. Then add the square of -4i to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-8iz+\left(-4i\right)^{2}=-20+\left(-4i\right)^{2}$$

Square -4i.

$$z^{2}-8iz-16=-20-16$$

Add -20 to -16.

$$z^{2}-8iz-16=-36$$

Factor z^{2}-8iz-16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-4i\right)^{2}=-36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-4i\right)^{2}}=\sqrt{-36}$$

Simplify.

$$z-4i=6i z-4i=-6i$$

Add 4i to both sides of the equation.

$$z=10i z=-2i$$