Solve 3/6+2i

$$\frac { 3 } { 6 + 2 i }$$

Evaluate

$$\frac{9}{20}-\frac{3}{20}i=0.45-0.15i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 6-2i.

$$\frac{3\left(6-2i\right)}{\left(6+2i\right)\left(6-2i\right)}$$

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

$$\frac{3\left(6-2i\right)}{6^{2}-2^{2}i^{2}}$$

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

$$\frac{3\left(6-2i\right)}{40}$$

Multiply 3 times 6-2i.

$$\frac{3\times 6+3\times \left(-2i\right)}{40}$$

Do the multiplications in 3\times 6+3\times \left(-2i\right).

$$\frac{18-6i}{40}$$

Divide 18-6i by 40 to get \frac{9}{20}-\frac{3}{20}i.

$$\frac{9}{20}-\frac{3}{20}i$$

Real Part

$$\frac{9}{20} = 0.45$$

Solution Steps

Multiply both numerator and denominator of \frac{3}{6+2i} by the complex conjugate of the denominator, 6-2i.

$$Re(\frac{3\left(6-2i\right)}{\left(6+2i\right)\left(6-2i\right)})$$

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

$$Re(\frac{3\left(6-2i\right)}{6^{2}-2^{2}i^{2}})$$

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

$$Re(\frac{3\left(6-2i\right)}{40})$$

Multiply 3 times 6-2i.

$$Re(\frac{3\times 6+3\times \left(-2i\right)}{40})$$

Do the multiplications in 3\times 6+3\times \left(-2i\right).

$$Re(\frac{18-6i}{40})$$

Divide 18-6i by 40 to get \frac{9}{20}-\frac{3}{20}i.

$$Re(\frac{9}{20}-\frac{3}{20}i)$$

The real part of \frac{9}{20}-\frac{3}{20}i is \frac{9}{20}.

$$\frac{9}{20}$$