Solve -4i/-5-4i

$$\frac { - 4 i } { - 5 - 4 i }$$

Evaluate

$$\frac{16}{41}+\frac{20}{41}i\approx 0.390243902+0.487804878i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -5+4i.

$$\frac{-4i\left(-5+4i\right)}{\left(-5-4i\right)\left(-5+4i\right)}$$

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

$$\frac{-4i\left(-5+4i\right)}{\left(-5\right)^{2}-4^{2}i^{2}}$$

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

$$\frac{-4i\left(-5+4i\right)}{41}$$

Multiply -4i times -5+4i.

$$\frac{-4i\left(-5\right)-4\times 4i^{2}}{41}$$

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

$$\frac{-4i\left(-5\right)-4\times 4\left(-1\right)}{41}$$

Do the multiplications in -4i\left(-5\right)-4\times 4\left(-1\right). Reorder the terms.

$$\frac{16+20i}{41}$$

Divide 16+20i by 41 to get \frac{16}{41}+\frac{20}{41}i.

$$\frac{16}{41}+\frac{20}{41}i$$

Real Part

$$\frac{16}{41} = 0.3902439024390244$$

Solution Steps

Multiply both numerator and denominator of \frac{-4i}{-5-4i} by the complex conjugate of the denominator, -5+4i.

$$Re(\frac{-4i\left(-5+4i\right)}{\left(-5-4i\right)\left(-5+4i\right)})$$

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

$$Re(\frac{-4i\left(-5+4i\right)}{\left(-5\right)^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{-4i\left(-5+4i\right)}{41})$$

Multiply -4i times -5+4i.

$$Re(\frac{-4i\left(-5\right)-4\times 4i^{2}}{41})$$

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

$$Re(\frac{-4i\left(-5\right)-4\times 4\left(-1\right)}{41})$$

Do the multiplications in -4i\left(-5\right)-4\times 4\left(-1\right). Reorder the terms.

$$Re(\frac{16+20i}{41})$$

Divide 16+20i by 41 to get \frac{16}{41}+\frac{20}{41}i.

$$Re(\frac{16}{41}+\frac{20}{41}i)$$

The real part of \frac{16}{41}+\frac{20}{41}i is \frac{16}{41}.

$$\frac{16}{41}$$