Solve z=1+2i/1+3i

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

Solve for z

$$z=\frac{7}{10}-\frac{1}{10}i=0.7-0.1i$$

Steps for Solving Linear Equation

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

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

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

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

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

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

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

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

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

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

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

$$z=\frac{1-3i+2i+6}{10}$$

Combine the real and imaginary parts in 1-3i+2i+6.

$$z=\frac{1+6+\left(-3+2\right)i}{10}$$

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

$$z=\frac{7-i}{10}$$

Divide 7-i by 10 to get \frac{7}{10}-\frac{1}{10}i.

$$z=\frac{7}{10}-\frac{1}{10}i$$

Assign z

$$z≔\frac{7}{10}-\frac{1}{10}i$$