Solve 6i/3-5i

$$\frac { 6 i } { 3 - 5 i }$$

Evaluate

$$-\frac{15}{17}+\frac{9}{17}i\approx -0.882352941+0.529411765i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+5i.

$$\frac{6i\left(3+5i\right)}{\left(3-5i\right)\left(3+5i\right)}$$

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

$$\frac{6i\left(3+5i\right)}{3^{2}-5^{2}i^{2}}$$

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

$$\frac{6i\left(3+5i\right)}{34}$$

Multiply 6i times 3+5i.

$$\frac{6i\times 3+6\times 5i^{2}}{34}$$

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

$$\frac{6i\times 3+6\times 5\left(-1\right)}{34}$$

Do the multiplications in 6i\times 3+6\times 5\left(-1\right). Reorder the terms.

$$\frac{-30+18i}{34}$$

Divide -30+18i by 34 to get -\frac{15}{17}+\frac{9}{17}i.

$$-\frac{15}{17}+\frac{9}{17}i$$

Real Part

$$-\frac{15}{17} = -0.8823529411764706$$

Solution Steps

Multiply both numerator and denominator of \frac{6i}{3-5i} by the complex conjugate of the denominator, 3+5i.

$$Re(\frac{6i\left(3+5i\right)}{\left(3-5i\right)\left(3+5i\right)})$$

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

$$Re(\frac{6i\left(3+5i\right)}{3^{2}-5^{2}i^{2}})$$

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

$$Re(\frac{6i\left(3+5i\right)}{34})$$

Multiply 6i times 3+5i.

$$Re(\frac{6i\times 3+6\times 5i^{2}}{34})$$

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

$$Re(\frac{6i\times 3+6\times 5\left(-1\right)}{34})$$

Do the multiplications in 6i\times 3+6\times 5\left(-1\right). Reorder the terms.

$$Re(\frac{-30+18i}{34})$$

Divide -30+18i by 34 to get -\frac{15}{17}+\frac{9}{17}i.

$$Re(-\frac{15}{17}+\frac{9}{17}i)$$

The real part of -\frac{15}{17}+\frac{9}{17}i is -\frac{15}{17}.

$$-\frac{15}{17}$$