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

$$2 - 2 z ( 1 + i ) = 4 i - 2$$

Solve for z

$$z=-2i$$

Steps for Solving Linear Equation

Multiply 2 times 1+i.

$$2-\left(2\times 1+2i\right)z=4i-2$$

Do the multiplications in 2\times 1+2i.

$$2-\left(2+2i\right)z=4i-2$$

Multiply -1 and 2+2i to get -2-2i.

$$2+\left(-2-2i\right)z=4i-2$$

Subtract 2 from both sides.

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

Combine the real and imaginary parts in 4i-2-2.

$$\left(-2-2i\right)z=-2-2+4i$$

Add -2 to -2.

$$\left(-2-2i\right)z=-4+4i$$

Divide both sides by -2-2i.

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

Multiply both numerator and denominator of \frac{-4+4i}{-2-2i} by the complex conjugate of the denominator, -2+2i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply complex numbers -4+4i and -2+2i like you multiply binomials.

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

By definition, i^{2} is -1.

$$z=\frac{-4\left(-2\right)-4\times \left(2i\right)+4i\left(-2\right)+4\times 2\left(-1\right)}{8}$$

Do the multiplications in -4\left(-2\right)-4\times \left(2i\right)+4i\left(-2\right)+4\times 2\left(-1\right).

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

Combine the real and imaginary parts in 8-8i-8i-8.

$$z=\frac{8-8+\left(-8-8\right)i}{8}$$

Do the additions in 8-8+\left(-8-8\right)i.

$$z=\frac{-16i}{8}$$

Divide -16i by 8 to get -2i.

$$z=-2i$$