Solve 2/3+2i

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

Evaluate

$$\frac{6}{13}-\frac{4}{13}i\approx 0.461538462-0.307692308i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-2i.

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

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

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

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

$$\frac{2\left(3-2i\right)}{13}$$

Multiply 2 times 3-2i.

$$\frac{2\times 3+2\times \left(-2i\right)}{13}$$

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

$$\frac{6-4i}{13}$$

Divide 6-4i by 13 to get \frac{6}{13}-\frac{4}{13}i.

$$\frac{6}{13}-\frac{4}{13}i$$

Real Part

$$\frac{6}{13} = 0.46153846153846156$$

Solution Steps

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

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

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

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

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

$$Re(\frac{2\left(3-2i\right)}{13})$$

Multiply 2 times 3-2i.

$$Re(\frac{2\times 3+2\times \left(-2i\right)}{13})$$

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

$$Re(\frac{6-4i}{13})$$

Divide 6-4i by 13 to get \frac{6}{13}-\frac{4}{13}i.

$$Re(\frac{6}{13}-\frac{4}{13}i)$$

The real part of \frac{6}{13}-\frac{4}{13}i is \frac{6}{13}.

$$\frac{6}{13}$$