Solve i/-2-8i

$$\frac { i } { - 2 - 8 i }$$

Evaluate

$$-\frac{2}{17}-\frac{1}{34}i\approx -0.117647059-0.029411765i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -2+8i.

$$\frac{i\left(-2+8i\right)}{\left(-2-8i\right)\left(-2+8i\right)}$$

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

$$\frac{i\left(-2+8i\right)}{\left(-2\right)^{2}-8^{2}i^{2}}$$

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

$$\frac{i\left(-2+8i\right)}{68}$$

Multiply i times -2+8i.

$$\frac{-2i+8i^{2}}{68}$$

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

$$\frac{-2i+8\left(-1\right)}{68}$$

Do the multiplications in -2i+8\left(-1\right). Reorder the terms.

$$\frac{-8-2i}{68}$$

Divide -8-2i by 68 to get -\frac{2}{17}-\frac{1}{34}i.

$$-\frac{2}{17}-\frac{1}{34}i$$

Real Part

$$-\frac{2}{17} = -0.11764705882352941$$

Solution Steps

Multiply both numerator and denominator of \frac{i}{-2-8i} by the complex conjugate of the denominator, -2+8i.

$$Re(\frac{i\left(-2+8i\right)}{\left(-2-8i\right)\left(-2+8i\right)})$$

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

$$Re(\frac{i\left(-2+8i\right)}{\left(-2\right)^{2}-8^{2}i^{2}})$$

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

$$Re(\frac{i\left(-2+8i\right)}{68})$$

Multiply i times -2+8i.

$$Re(\frac{-2i+8i^{2}}{68})$$

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

$$Re(\frac{-2i+8\left(-1\right)}{68})$$

Do the multiplications in -2i+8\left(-1\right). Reorder the terms.

$$Re(\frac{-8-2i}{68})$$

Divide -8-2i by 68 to get -\frac{2}{17}-\frac{1}{34}i.

$$Re(-\frac{2}{17}-\frac{1}{34}i)$$

The real part of -\frac{2}{17}-\frac{1}{34}i is -\frac{2}{17}.

$$-\frac{2}{17}$$