Solve 4i/5-3i

$$\frac { 4 i } { 5 - 3 i }$$

Evaluate

$$-\frac{6}{17}+\frac{10}{17}i\approx -0.352941176+0.588235294i$$

Solution Steps

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

$$\frac{4i\left(5+3i\right)}{\left(5-3i\right)\left(5+3i\right)}$$

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

$$\frac{4i\left(5+3i\right)}{5^{2}-3^{2}i^{2}}$$

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

$$\frac{4i\left(5+3i\right)}{34}$$

Multiply 4i times 5+3i.

$$\frac{4i\times 5+4\times 3i^{2}}{34}$$

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

$$\frac{4i\times 5+4\times 3\left(-1\right)}{34}$$

Do the multiplications in 4i\times 5+4\times 3\left(-1\right). Reorder the terms.

$$\frac{-12+20i}{34}$$

Divide -12+20i by 34 to get -\frac{6}{17}+\frac{10}{17}i.

$$-\frac{6}{17}+\frac{10}{17}i$$

Real Part

$$-\frac{6}{17} = -0.35294117647058826$$

Solution Steps

Multiply both numerator and denominator of \frac{4i}{5-3i} by the complex conjugate of the denominator, 5+3i.

$$Re(\frac{4i\left(5+3i\right)}{\left(5-3i\right)\left(5+3i\right)})$$

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

$$Re(\frac{4i\left(5+3i\right)}{5^{2}-3^{2}i^{2}})$$

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

$$Re(\frac{4i\left(5+3i\right)}{34})$$

Multiply 4i times 5+3i.

$$Re(\frac{4i\times 5+4\times 3i^{2}}{34})$$

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

$$Re(\frac{4i\times 5+4\times 3\left(-1\right)}{34})$$

Do the multiplications in 4i\times 5+4\times 3\left(-1\right). Reorder the terms.

$$Re(\frac{-12+20i}{34})$$

Divide -12+20i by 34 to get -\frac{6}{17}+\frac{10}{17}i.

$$Re(-\frac{6}{17}+\frac{10}{17}i)$$

The real part of -\frac{6}{17}+\frac{10}{17}i is -\frac{6}{17}.

$$-\frac{6}{17}$$