Solve 2i/4-3i

$$\frac { 2 i } { 4 - 3 i }$$

Evaluate

$$-\frac{6}{25}+\frac{8}{25}i=-0.24+0.32i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4+3i.

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

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

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

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

$$\frac{2i\left(4+3i\right)}{25}$$

Multiply 2i times 4+3i.

$$\frac{2i\times 4+2\times 3i^{2}}{25}$$

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

$$\frac{2i\times 4+2\times 3\left(-1\right)}{25}$$

Do the multiplications in 2i\times 4+2\times 3\left(-1\right). Reorder the terms.

$$\frac{-6+8i}{25}$$

Divide -6+8i by 25 to get -\frac{6}{25}+\frac{8}{25}i.

$$-\frac{6}{25}+\frac{8}{25}i$$

Real Part

$$-\frac{6}{25} = -0.24$$

Solution Steps

Multiply both numerator and denominator of \frac{2i}{4-3i} by the complex conjugate of the denominator, 4+3i.

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

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

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

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

$$Re(\frac{2i\left(4+3i\right)}{25})$$

Multiply 2i times 4+3i.

$$Re(\frac{2i\times 4+2\times 3i^{2}}{25})$$

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

$$Re(\frac{2i\times 4+2\times 3\left(-1\right)}{25})$$

Do the multiplications in 2i\times 4+2\times 3\left(-1\right). Reorder the terms.

$$Re(\frac{-6+8i}{25})$$

Divide -6+8i by 25 to get -\frac{6}{25}+\frac{8}{25}i.

$$Re(-\frac{6}{25}+\frac{8}{25}i)$$

The real part of -\frac{6}{25}+\frac{8}{25}i is -\frac{6}{25}.

$$-\frac{6}{25}$$