Solve 3-6i/18i

$$\frac { 3 - 6 i } { 18 i }$$

Evaluate

$$-\frac{1}{3}-\frac{1}{6}i\approx -0.333333333-0.166666667i$$

Solution Steps

Multiply both numerator and denominator by imaginary unit i.

$$\frac{\left(3-6i\right)i}{18i^{2}}$$

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

$$\frac{\left(3-6i\right)i}{-18}$$

Multiply 3-6i times i.

$$\frac{3i-6i^{2}}{-18}$$

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

$$\frac{3i-6\left(-1\right)}{-18}$$

Do the multiplications in 3i-6\left(-1\right). Reorder the terms.

$$\frac{6+3i}{-18}$$

Divide 6+3i by -18 to get -\frac{1}{3}-\frac{1}{6}i.

$$-\frac{1}{3}-\frac{1}{6}i$$

Real Part

$$-\frac{1}{3} = -0.3333333333333333$$

Solution Steps

Multiply both numerator and denominator of \frac{3-6i}{18i} by imaginary unit i.

$$Re(\frac{\left(3-6i\right)i}{18i^{2}})$$

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

$$Re(\frac{\left(3-6i\right)i}{-18})$$

Multiply 3-6i times i.

$$Re(\frac{3i-6i^{2}}{-18})$$

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

$$Re(\frac{3i-6\left(-1\right)}{-18})$$

Do the multiplications in 3i-6\left(-1\right). Reorder the terms.

$$Re(\frac{6+3i}{-18})$$

Divide 6+3i by -18 to get -\frac{1}{3}-\frac{1}{6}i.

$$Re(-\frac{1}{3}-\frac{1}{6}i)$$

The real part of -\frac{1}{3}-\frac{1}{6}i is -\frac{1}{3}.

$$-\frac{1}{3}$$