Solve 4i/-1+4i

$$\frac { 4 i } { - 1 + 4 i }$$

Evaluate

$$\frac{16}{17}-\frac{4}{17}i\approx 0.941176471-0.235294118i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -1-4i.

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

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

$$\frac{4i\left(-1-4i\right)}{\left(-1\right)^{2}-4^{2}i^{2}}$$

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

$$\frac{4i\left(-1-4i\right)}{17}$$

Multiply 4i times -1-4i.

$$\frac{4i\left(-1\right)+4\left(-4\right)i^{2}}{17}$$

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

$$\frac{4i\left(-1\right)+4\left(-4\right)\left(-1\right)}{17}$$

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

$$\frac{16-4i}{17}$$

Divide 16-4i by 17 to get \frac{16}{17}-\frac{4}{17}i.

$$\frac{16}{17}-\frac{4}{17}i$$

Real Part

$$\frac{16}{17} = 0.9411764705882353$$

Solution Steps

Multiply both numerator and denominator of \frac{4i}{-1+4i} by the complex conjugate of the denominator, -1-4i.

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

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

$$Re(\frac{4i\left(-1-4i\right)}{\left(-1\right)^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{4i\left(-1-4i\right)}{17})$$

Multiply 4i times -1-4i.

$$Re(\frac{4i\left(-1\right)+4\left(-4\right)i^{2}}{17})$$

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

$$Re(\frac{4i\left(-1\right)+4\left(-4\right)\left(-1\right)}{17})$$

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

$$Re(\frac{16-4i}{17})$$

Divide 16-4i by 17 to get \frac{16}{17}-\frac{4}{17}i.

$$Re(\frac{16}{17}-\frac{4}{17}i)$$

The real part of \frac{16}{17}-\frac{4}{17}i is \frac{16}{17}.

$$\frac{16}{17}$$