Solve 6i/4-5i

$$\frac { 6 i } { 4 - 5 i }$$

Evaluate

$$-\frac{30}{41}+\frac{24}{41}i\approx -0.731707317+0.585365854i$$

Solution Steps

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

$$\frac{6i\left(4+5i\right)}{\left(4-5i\right)\left(4+5i\right)}$$

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

$$\frac{6i\left(4+5i\right)}{4^{2}-5^{2}i^{2}}$$

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

$$\frac{6i\left(4+5i\right)}{41}$$

Multiply 6i times 4+5i.

$$\frac{6i\times 4+6\times 5i^{2}}{41}$$

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

$$\frac{6i\times 4+6\times 5\left(-1\right)}{41}$$

Do the multiplications in 6i\times 4+6\times 5\left(-1\right). Reorder the terms.

$$\frac{-30+24i}{41}$$

Divide -30+24i by 41 to get -\frac{30}{41}+\frac{24}{41}i.

$$-\frac{30}{41}+\frac{24}{41}i$$

Real Part

$$-\frac{30}{41} = -0.7317073170731707$$

Solution Steps

Multiply both numerator and denominator of \frac{6i}{4-5i} by the complex conjugate of the denominator, 4+5i.

$$Re(\frac{6i\left(4+5i\right)}{\left(4-5i\right)\left(4+5i\right)})$$

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

$$Re(\frac{6i\left(4+5i\right)}{4^{2}-5^{2}i^{2}})$$

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

$$Re(\frac{6i\left(4+5i\right)}{41})$$

Multiply 6i times 4+5i.

$$Re(\frac{6i\times 4+6\times 5i^{2}}{41})$$

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

$$Re(\frac{6i\times 4+6\times 5\left(-1\right)}{41})$$

Do the multiplications in 6i\times 4+6\times 5\left(-1\right). Reorder the terms.

$$Re(\frac{-30+24i}{41})$$

Divide -30+24i by 41 to get -\frac{30}{41}+\frac{24}{41}i.

$$Re(-\frac{30}{41}+\frac{24}{41}i)$$

The real part of -\frac{30}{41}+\frac{24}{41}i is -\frac{30}{41}.

$$-\frac{30}{41}$$