Solve 3i/(5+4i)

$$3 i / ( 5 + 4 i )$$

Evaluate

$$\frac{12}{41}+\frac{15}{41}i\approx 0.292682927+0.365853659i$$

Solution Steps

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

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

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

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

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

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

Multiply 3i times 5-4i.

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

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

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

Do the multiplications in 3i\times 5+3\left(-4\right)\left(-1\right). Reorder the terms.

$$\frac{12+15i}{41}$$

Divide 12+15i by 41 to get \frac{12}{41}+\frac{15}{41}i.

$$\frac{12}{41}+\frac{15}{41}i$$

Real Part

$$\frac{12}{41} = 0.2926829268292683$$

Solution Steps

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

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

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

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

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

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

Multiply 3i times 5-4i.

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

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

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

Do the multiplications in 3i\times 5+3\left(-4\right)\left(-1\right). Reorder the terms.

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

Divide 12+15i by 41 to get \frac{12}{41}+\frac{15}{41}i.

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

The real part of \frac{12}{41}+\frac{15}{41}i is \frac{12}{41}.

$$\frac{12}{41}$$