Solve z=1+i/2-i-i

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

Solve for z

$$z=\frac{1}{5}-\frac{2}{5}i=0.2-0.4i$$

Steps for Solving Linear Equation

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

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

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

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

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

$$z=\frac{\left(1+i\right)\left(2+i\right)}{5}-i$$

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

$$z=\frac{1\times 2+i+2i+i^{2}}{5}-i$$

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

$$z=\frac{1\times 2+i+2i-1}{5}-i$$

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

$$z=\frac{2+i+2i-1}{5}-i$$

Combine the real and imaginary parts in 2+i+2i-1.

$$z=\frac{2-1+\left(1+2\right)i}{5}-i$$

Do the additions in 2-1+\left(1+2\right)i.

$$z=\frac{1+3i}{5}-i$$

Divide 1+3i by 5 to get \frac{1}{5}+\frac{3}{5}i.

$$z=\frac{1}{5}+\frac{3}{5}i-i$$

Combine the real and imaginary parts in numbers \frac{1}{5}+\frac{3}{5}i and -i.

$$z=\frac{1}{5}+\left(\frac{3}{5}-1\right)i$$

Add \frac{3}{5} to -1.

$$z=\frac{1}{5}-\frac{2}{5}i$$

Assign z

$$z≔\frac{1}{5}-\frac{2}{5}i$$