Solve z=1-i/3+i

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

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}{3+i} by the complex conjugate of the denominator, 3-i.

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

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

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

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

$$z=\frac{\left(1-i\right)\left(3-i\right)}{10}$$

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

$$z=\frac{1\times 3+1\left(-i\right)-i\times 3-\left(-i^{2}\right)}{10}$$

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

$$z=\frac{1\times 3+1\left(-i\right)-i\times 3-\left(-\left(-1\right)\right)}{10}$$

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

$$z=\frac{3-i-3i-1}{10}$$

Combine the real and imaginary parts in 3-i-3i-1.

$$z=\frac{3-1+\left(-1-3\right)i}{10}$$

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

$$z=\frac{2-4i}{10}$$

Divide 2-4i by 10 to get \frac{1}{5}-\frac{2}{5}i.

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

Assign z

$$z≔\frac{1}{5}-\frac{2}{5}i$$