Solve 3/2+4i

$$\frac { 3 } { 2 + 4 i }$$

Evaluate

$$\frac{3}{10}-\frac{3}{5}i=0.3-0.6i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-4i.

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

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

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

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

$$\frac{3\left(2-4i\right)}{20}$$

Multiply 3 times 2-4i.

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

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

$$\frac{6-12i}{20}$$

Divide 6-12i by 20 to get \frac{3}{10}-\frac{3}{5}i.

$$\frac{3}{10}-\frac{3}{5}i$$

Real Part

$$\frac{3}{10} = 0.3$$

Solution Steps

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

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

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

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

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

$$Re(\frac{3\left(2-4i\right)}{20})$$

Multiply 3 times 2-4i.

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

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

$$Re(\frac{6-12i}{20})$$

Divide 6-12i by 20 to get \frac{3}{10}-\frac{3}{5}i.

$$Re(\frac{3}{10}-\frac{3}{5}i)$$

The real part of \frac{3}{10}-\frac{3}{5}i is \frac{3}{10}.

$$\frac{3}{10}$$