Solve i/2+3i

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

Evaluate

$$\frac{3}{13}+\frac{2}{13}i\approx 0.230769231+0.153846154i$$

Solution Steps

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

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

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

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

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

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

Multiply i times 2-3i.

$$\frac{2i-3i^{2}}{13}$$

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

$$\frac{2i-3\left(-1\right)}{13}$$

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

$$\frac{3+2i}{13}$$

Divide 3+2i by 13 to get \frac{3}{13}+\frac{2}{13}i.

$$\frac{3}{13}+\frac{2}{13}i$$

Real Part

$$\frac{3}{13} = 0.23076923076923078$$

Solution Steps

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

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

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

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

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

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

Multiply i times 2-3i.

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

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

$$Re(\frac{2i-3\left(-1\right)}{13})$$

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

$$Re(\frac{3+2i}{13})$$

Divide 3+2i by 13 to get \frac{3}{13}+\frac{2}{13}i.

$$Re(\frac{3}{13}+\frac{2}{13}i)$$

The real part of \frac{3}{13}+\frac{2}{13}i is \frac{3}{13}.

$$\frac{3}{13}$$