Solve i(3+4i)/(3-4i)(3+4i)

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

Evaluate

$$-\frac{4}{25}+\frac{3}{25}i=-0.16+0.12i$$

Solution Steps

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

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

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

$$\frac{i\left(3+4i\right)}{25}$$

Multiply i times 3+4i.

$$\frac{3i+4i^{2}}{25}$$

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

$$\frac{3i+4\left(-1\right)}{25}$$

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

$$\frac{-4+3i}{25}$$

Divide -4+3i by 25 to get -\frac{4}{25}+\frac{3}{25}i.

$$-\frac{4}{25}+\frac{3}{25}i$$

Real Part

$$-\frac{4}{25} = -0.16$$

Solution Steps

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

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

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

$$Re(\frac{i\left(3+4i\right)}{25})$$

Multiply i times 3+4i.

$$Re(\frac{3i+4i^{2}}{25})$$

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

$$Re(\frac{3i+4\left(-1\right)}{25})$$

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

$$Re(\frac{-4+3i}{25})$$

Divide -4+3i by 25 to get -\frac{4}{25}+\frac{3}{25}i.

$$Re(-\frac{4}{25}+\frac{3}{25}i)$$

The real part of -\frac{4}{25}+\frac{3}{25}i is -\frac{4}{25}.

$$-\frac{4}{25}$$