Solve +yi=2+3i/4-2i

$$+ y i = \frac { 2 + 3 i } { 4 - 2 i }$$

Solve for y

$$y=\frac{4}{5}-\frac{1}{10}i=0.8-0.1i$$

Steps for Solving Linear Equation

Multiply both numerator and denominator of \frac{2+3i}{4-2i} by the complex conjugate of the denominator, 4+2i.

$$yi=\frac{\left(2+3i\right)\left(4+2i\right)}{\left(4-2i\right)\left(4+2i\right)}$$

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

$$yi=\frac{\left(2+3i\right)\left(4+2i\right)}{4^{2}-2^{2}i^{2}}$$

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

$$yi=\frac{\left(2+3i\right)\left(4+2i\right)}{20}$$

Multiply complex numbers 2+3i and 4+2i like you multiply binomials.

$$yi=\frac{2\times 4+2\times \left(2i\right)+3i\times 4+3\times 2i^{2}}{20}$$

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

$$yi=\frac{2\times 4+2\times \left(2i\right)+3i\times 4+3\times 2\left(-1\right)}{20}$$

Do the multiplications in 2\times 4+2\times \left(2i\right)+3i\times 4+3\times 2\left(-1\right).

$$yi=\frac{8+4i+12i-6}{20}$$

Combine the real and imaginary parts in 8+4i+12i-6.

$$yi=\frac{8-6+\left(4+12\right)i}{20}$$

Do the additions in 8-6+\left(4+12\right)i.

$$yi=\frac{2+16i}{20}$$

Divide 2+16i by 20 to get \frac{1}{10}+\frac{4}{5}i.

$$yi=\frac{1}{10}+\frac{4}{5}i$$

Divide both sides by i.

$$y=\frac{\frac{1}{10}+\frac{4}{5}i}{i}$$

Multiply both numerator and denominator of \frac{\frac{1}{10}+\frac{4}{5}i}{i} by imaginary unit i.

$$y=\frac{\left(\frac{1}{10}+\frac{4}{5}i\right)i}{1i^{2}}$$

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

$$y=\frac{\left(\frac{1}{10}+\frac{4}{5}i\right)i}{-1}$$

Multiply \frac{1}{10}+\frac{4}{5}i times i.

$$y=\frac{\frac{1}{10}i+\frac{4}{5}i^{2}}{-1}$$

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

$$y=\frac{\frac{1}{10}i+\frac{4}{5}\left(-1\right)}{-1}$$

Do the multiplications in \frac{1}{10}i+\frac{4}{5}\left(-1\right). Reorder the terms.

$$y=\frac{-\frac{4}{5}+\frac{1}{10}i}{-1}$$

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

$$y=\frac{4}{5}-\frac{1}{10}i$$