Solve z=i/-2-2i

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

Solve for z

$$z=-\frac{1}{4}-\frac{1}{4}i=-0.25-0.25i$$

Steps for Solving Linear Equation

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

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

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

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

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

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

Multiply i times -2+2i.

$$z=\frac{-2i+2i^{2}}{8}$$

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

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

Do the multiplications in -2i+2\left(-1\right). Reorder the terms.

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

Divide -2-2i by 8 to get -\frac{1}{4}-\frac{1}{4}i.

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

Assign z

$$z≔-\frac{1}{4}-\frac{1}{4}i$$