Solve z=1+3i/2-i(i)

$$z = \frac { 1 + 3 i } { 2 - i } ( i )$$

Solve for z

$$z=-\frac{7}{5}-\frac{1}{5}i=-1.4-0.2i$$

Steps for Solving Linear Equation

Multiply both numerator and denominator of \frac{1+3i}{2-i} by the complex conjugate of the denominator, 2+i.

$$z=\frac{\left(1+3i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}i$$

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

$$z=\frac{\left(1+3i\right)\left(2+i\right)}{2^{2}-i^{2}}i$$

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

$$z=\frac{\left(1+3i\right)\left(2+i\right)}{5}i$$

Multiply complex numbers 1+3i and 2+i like you multiply binomials.

$$z=\frac{1\times 2+i+3i\times 2+3i^{2}}{5}i$$

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

$$z=\frac{1\times 2+i+3i\times 2+3\left(-1\right)}{5}i$$

Do the multiplications in 1\times 2+i+3i\times 2+3\left(-1\right).

$$z=\frac{2+i+6i-3}{5}i$$

Combine the real and imaginary parts in 2+i+6i-3.

$$z=\frac{2-3+\left(1+6\right)i}{5}i$$

Do the additions in 2-3+\left(1+6\right)i.

$$z=\frac{-1+7i}{5}i$$

Divide -1+7i by 5 to get -\frac{1}{5}+\frac{7}{5}i.

$$z=\left(-\frac{1}{5}+\frac{7}{5}i\right)i$$

Multiply -\frac{1}{5}+\frac{7}{5}i times i.

$$z=-\frac{1}{5}i+\frac{7}{5}i^{2}$$

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

$$z=-\frac{1}{5}i+\frac{7}{5}\left(-1\right)$$

Do the multiplications in -\frac{1}{5}i+\frac{7}{5}\left(-1\right). Reorder the terms.

$$z=-\frac{7}{5}-\frac{1}{5}i$$

Assign z

$$z≔-\frac{7}{5}-\frac{1}{5}i$$