Solve z+zi=4-2i

$$z + z i = 4 - 2 i$$

Solve for z

$$z=1-3i$$

Steps for Solving Linear Equation

Combine z and zi to get \left(1+i\right)z.

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

Divide both sides by 1+i.

$$z=\frac{4-2i}{1+i}$$

Multiply both numerator and denominator of \frac{4-2i}{1+i} by the complex conjugate of the denominator, 1-i.

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

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

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

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

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

Multiply complex numbers 4-2i and 1-i like you multiply binomials.

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

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

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

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

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

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

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

Do the additions in 4-2+\left(-4-2\right)i.

$$z=\frac{2-6i}{2}$$

Divide 2-6i by 2 to get 1-3i.

$$z=1-3i$$