Solve 2i/9-6i

$$\frac { 2 i } { 9 - 6 i }$$

Evaluate

$$-\frac{4}{39}+\frac{2}{13}i\approx -0.102564103+0.153846154i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 9+6i.

$$\frac{2i\left(9+6i\right)}{\left(9-6i\right)\left(9+6i\right)}$$

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

$$\frac{2i\left(9+6i\right)}{9^{2}-6^{2}i^{2}}$$

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

$$\frac{2i\left(9+6i\right)}{117}$$

Multiply 2i times 9+6i.

$$\frac{2i\times 9+2\times 6i^{2}}{117}$$

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

$$\frac{2i\times 9+2\times 6\left(-1\right)}{117}$$

Do the multiplications in 2i\times 9+2\times 6\left(-1\right). Reorder the terms.

$$\frac{-12+18i}{117}$$

Divide -12+18i by 117 to get -\frac{4}{39}+\frac{2}{13}i.

$$-\frac{4}{39}+\frac{2}{13}i$$

Real Part

$$-\frac{4}{39} = -0.10256410256410256$$

Solution Steps

Multiply both numerator and denominator of \frac{2i}{9-6i} by the complex conjugate of the denominator, 9+6i.

$$Re(\frac{2i\left(9+6i\right)}{\left(9-6i\right)\left(9+6i\right)})$$

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

$$Re(\frac{2i\left(9+6i\right)}{9^{2}-6^{2}i^{2}})$$

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

$$Re(\frac{2i\left(9+6i\right)}{117})$$

Multiply 2i times 9+6i.

$$Re(\frac{2i\times 9+2\times 6i^{2}}{117})$$

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

$$Re(\frac{2i\times 9+2\times 6\left(-1\right)}{117})$$

Do the multiplications in 2i\times 9+2\times 6\left(-1\right). Reorder the terms.

$$Re(\frac{-12+18i}{117})$$

Divide -12+18i by 117 to get -\frac{4}{39}+\frac{2}{13}i.

$$Re(-\frac{4}{39}+\frac{2}{13}i)$$

The real part of -\frac{4}{39}+\frac{2}{13}i is -\frac{4}{39}.

$$-\frac{4}{39}$$