Solve 15/3+4i

$$\frac { 15 } { 3 + 4 i }$$

Evaluate

$$\frac{9}{5}-\frac{12}{5}i=1.8-2.4i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-4i.

$$\frac{15\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}$$

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

$$\frac{15\left(3-4i\right)}{3^{2}-4^{2}i^{2}}$$

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

$$\frac{15\left(3-4i\right)}{25}$$

Multiply 15 times 3-4i.

$$\frac{15\times 3+15\times \left(-4i\right)}{25}$$

Do the multiplications in 15\times 3+15\times \left(-4i\right).

$$\frac{45-60i}{25}$$

Divide 45-60i by 25 to get \frac{9}{5}-\frac{12}{5}i.

$$\frac{9}{5}-\frac{12}{5}i$$

Real Part

$$\frac{9}{5} = 1\frac{4}{5} = 1.8$$

Solution Steps

Multiply both numerator and denominator of \frac{15}{3+4i} by the complex conjugate of the denominator, 3-4i.

$$Re(\frac{15\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)})$$

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

$$Re(\frac{15\left(3-4i\right)}{3^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{15\left(3-4i\right)}{25})$$

Multiply 15 times 3-4i.

$$Re(\frac{15\times 3+15\times \left(-4i\right)}{25})$$

Do the multiplications in 15\times 3+15\times \left(-4i\right).

$$Re(\frac{45-60i}{25})$$

Divide 45-60i by 25 to get \frac{9}{5}-\frac{12}{5}i.

$$Re(\frac{9}{5}-\frac{12}{5}i)$$

The real part of \frac{9}{5}-\frac{12}{5}i is \frac{9}{5}.

$$\frac{9}{5}$$