Solve -12/5+3i/5+4i

$$\frac { - \frac { 12 } { 5 } + 3 i } { 5 + 4 i }$$

Evaluate

$$\frac{3}{5}i=0.6i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5-4i.

$$\frac{\left(-\frac{12}{5}+3i\right)\left(5-4i\right)}{\left(5+4i\right)\left(5-4i\right)}$$

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

$$\frac{\left(-\frac{12}{5}+3i\right)\left(5-4i\right)}{5^{2}-4^{2}i^{2}}$$

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

$$\frac{\left(-\frac{12}{5}+3i\right)\left(5-4i\right)}{41}$$

Multiply complex numbers -\frac{12}{5}+3i and 5-4i like you multiply binomials.

$$\frac{-\frac{12}{5}\times 5-\frac{12}{5}\times \left(-4i\right)+3i\times 5+3\left(-4\right)i^{2}}{41}$$

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

$$\frac{-\frac{12}{5}\times 5-\frac{12}{5}\times \left(-4i\right)+3i\times 5+3\left(-4\right)\left(-1\right)}{41}$$

Do the multiplications in -\frac{12}{5}\times 5-\frac{12}{5}\times \left(-4i\right)+3i\times 5+3\left(-4\right)\left(-1\right).

$$\frac{-12+\frac{48}{5}i+15i+12}{41}$$

Combine the real and imaginary parts in -12+\frac{48}{5}i+15i+12.

$$\frac{-12+12+\left(\frac{48}{5}+15\right)i}{41}$$

Do the additions in -12+12+\left(\frac{48}{5}+15\right)i.

$$\frac{\frac{123}{5}i}{41}$$

Divide \frac{123}{5}i by 41 to get \frac{3}{5}i.

$$\frac{3}{5}i$$

Real Part

$$0$$

Solution Steps

Multiply both numerator and denominator of \frac{-\frac{12}{5}+3i}{5+4i} by the complex conjugate of the denominator, 5-4i.

$$Re(\frac{\left(-\frac{12}{5}+3i\right)\left(5-4i\right)}{\left(5+4i\right)\left(5-4i\right)})$$

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

$$Re(\frac{\left(-\frac{12}{5}+3i\right)\left(5-4i\right)}{5^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{\left(-\frac{12}{5}+3i\right)\left(5-4i\right)}{41})$$

Multiply complex numbers -\frac{12}{5}+3i and 5-4i like you multiply binomials.

$$Re(\frac{-\frac{12}{5}\times 5-\frac{12}{5}\times \left(-4i\right)+3i\times 5+3\left(-4\right)i^{2}}{41})$$

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

$$Re(\frac{-\frac{12}{5}\times 5-\frac{12}{5}\times \left(-4i\right)+3i\times 5+3\left(-4\right)\left(-1\right)}{41})$$

Do the multiplications in -\frac{12}{5}\times 5-\frac{12}{5}\times \left(-4i\right)+3i\times 5+3\left(-4\right)\left(-1\right).

$$Re(\frac{-12+\frac{48}{5}i+15i+12}{41})$$

Combine the real and imaginary parts in -12+\frac{48}{5}i+15i+12.

$$Re(\frac{-12+12+\left(\frac{48}{5}+15\right)i}{41})$$

Do the additions in -12+12+\left(\frac{48}{5}+15\right)i.

$$Re(\frac{\frac{123}{5}i}{41})$$

Divide \frac{123}{5}i by 41 to get \frac{3}{5}i.

$$Re(\frac{3}{5}i)$$

The real part of \frac{3}{5}i is 0.

$$0$$