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

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

Solve for z

$$z=-\frac{3}{5}+\frac{1}{5}i=-0.6+0.2i$$

Steps for Solving Linear Equation

Use the distributive property to multiply i by i+1.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Assign z

$$z≔-\frac{3}{5}+\frac{1}{5}i$$