Solve c=(-3+2i):(2-i)

$$c = ( - 3 + 2 i ) : ( 2 - i )$$

Solve for c

$$c=-\frac{8}{5}+\frac{1}{5}i=-1.6+0.2i$$

Steps for Solving Linear Equation

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

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

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

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

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

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

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

$$c=\frac{-3\times 2-3i+2i\times 2+2i^{2}}{5}$$

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

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

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

$$c=\frac{-6-3i+4i-2}{5}$$

Combine the real and imaginary parts in -6-3i+4i-2.

$$c=\frac{-6-2+\left(-3+4\right)i}{5}$$

Do the additions in -6-2+\left(-3+4\right)i.

$$c=\frac{-8+i}{5}$$

Divide -8+i by 5 to get -\frac{8}{5}+\frac{1}{5}i.

$$c=-\frac{8}{5}+\frac{1}{5}i$$

Assign c

$$c≔-\frac{8}{5}+\frac{1}{5}i$$