Solve 3i/4+2i

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

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, 4-2i.

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

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

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

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

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

Multiply 3i times 4-2i.

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

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

$$\frac{3i\times 4+3\left(-2\right)\left(-1\right)}{20}$$

Do the multiplications in 3i\times 4+3\left(-2\right)\left(-1\right). Reorder the terms.

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

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{3i}{4+2i} by the complex conjugate of the denominator, 4-2i.

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

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

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

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

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

Multiply 3i times 4-2i.

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

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

$$Re(\frac{3i\times 4+3\left(-2\right)\left(-1\right)}{20})$$

Do the multiplications in 3i\times 4+3\left(-2\right)\left(-1\right). Reorder the terms.

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

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}.

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