Solve 5i/3-6i

$$\frac { 5 i } { 3 - 6 i }$$

Evaluate

$$-\frac{2}{3}+\frac{1}{3}i\approx -0.666666667+0.333333333i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+6i.

$$\frac{5i\left(3+6i\right)}{\left(3-6i\right)\left(3+6i\right)}$$

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

$$\frac{5i\left(3+6i\right)}{3^{2}-6^{2}i^{2}}$$

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

$$\frac{5i\left(3+6i\right)}{45}$$

Multiply 5i times 3+6i.

$$\frac{5i\times 3+5\times 6i^{2}}{45}$$

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

$$\frac{5i\times 3+5\times 6\left(-1\right)}{45}$$

Do the multiplications in 5i\times 3+5\times 6\left(-1\right). Reorder the terms.

$$\frac{-30+15i}{45}$$

Divide -30+15i by 45 to get -\frac{2}{3}+\frac{1}{3}i.

$$-\frac{2}{3}+\frac{1}{3}i$$

Real Part

$$-\frac{2}{3} = -0.6666666666666666$$

Solution Steps

Multiply both numerator and denominator of \frac{5i}{3-6i} by the complex conjugate of the denominator, 3+6i.

$$Re(\frac{5i\left(3+6i\right)}{\left(3-6i\right)\left(3+6i\right)})$$

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

$$Re(\frac{5i\left(3+6i\right)}{3^{2}-6^{2}i^{2}})$$

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

$$Re(\frac{5i\left(3+6i\right)}{45})$$

Multiply 5i times 3+6i.

$$Re(\frac{5i\times 3+5\times 6i^{2}}{45})$$

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

$$Re(\frac{5i\times 3+5\times 6\left(-1\right)}{45})$$

Do the multiplications in 5i\times 3+5\times 6\left(-1\right). Reorder the terms.

$$Re(\frac{-30+15i}{45})$$

Divide -30+15i by 45 to get -\frac{2}{3}+\frac{1}{3}i.

$$Re(-\frac{2}{3}+\frac{1}{3}i)$$

The real part of -\frac{2}{3}+\frac{1}{3}i is -\frac{2}{3}.

$$-\frac{2}{3}$$