Solve (z-2i)^2=(1-i)^4

$$( z - 2 i ) ^ { 2 } = ( 1 - i ) ^ { 4 }$$

Solve for z

$$z=0$$

$$z=4i$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z-2i\right)^{2}.

$$z^{2}-4iz-4=\left(1-i\right)^{4}$$

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

$$z^{2}-4iz-4=-4$$

Add 4 to both sides.

$$z^{2}-4iz-4+4=0$$

Add -4 and 4 to get 0.

$$z^{2}-4iz=0$$

Factor out z.

$$z\left(z-4i\right)=0$$

To find equation solutions, solve z=0 and z-4i=0.

$$z=0 z=4i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z-2i\right)^{2}.

$$z^{2}-4iz-4=\left(1-i\right)^{4}$$

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

$$z^{2}-4iz-4=-4$$

Add 4 to both sides.

$$z^{2}-4iz-4+4=0$$

Add -4 and 4 to get 0.

$$z^{2}-4iz=0$$

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

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

Take the square root of \left(-4i\right)^{2}.

$$z=\frac{4i±4i}{2}$$

Now solve the equation z=\frac{4i±4i}{2} when ± is plus. Add 4i to 4i.

$$z=\frac{8i}{2}$$

Divide 8i by 2.

$$z=4i$$

Now solve the equation z=\frac{4i±4i}{2} when ± is minus. Subtract 4i from 4i.

$$z=\frac{0}{2}$$

Divide 0 by 2.

$$z=0$$

The equation is now solved.

$$z=4i z=0$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z-2i\right)^{2}.

$$z^{2}-4iz-4=\left(1-i\right)^{4}$$

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

$$z^{2}-4iz-4=-4$$

Factor z^{2}-4iz-4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2i to both sides of the equation.

$$z=4i z=0$$