Solve 6i/7-3i

$$\frac { 6 i } { 7 - 3 i }$$

Evaluate

$$-\frac{9}{29}+\frac{21}{29}i\approx -0.310344828+0.724137931i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 7+3i.

$$\frac{6i\left(7+3i\right)}{\left(7-3i\right)\left(7+3i\right)}$$

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

$$\frac{6i\left(7+3i\right)}{7^{2}-3^{2}i^{2}}$$

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

$$\frac{6i\left(7+3i\right)}{58}$$

Multiply 6i times 7+3i.

$$\frac{6i\times 7+6\times 3i^{2}}{58}$$

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

$$\frac{6i\times 7+6\times 3\left(-1\right)}{58}$$

Do the multiplications in 6i\times 7+6\times 3\left(-1\right). Reorder the terms.

$$\frac{-18+42i}{58}$$

Divide -18+42i by 58 to get -\frac{9}{29}+\frac{21}{29}i.

$$-\frac{9}{29}+\frac{21}{29}i$$

Real Part

$$-\frac{9}{29} = -0.3103448275862069$$

Solution Steps

Multiply both numerator and denominator of \frac{6i}{7-3i} by the complex conjugate of the denominator, 7+3i.

$$Re(\frac{6i\left(7+3i\right)}{\left(7-3i\right)\left(7+3i\right)})$$

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

$$Re(\frac{6i\left(7+3i\right)}{7^{2}-3^{2}i^{2}})$$

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

$$Re(\frac{6i\left(7+3i\right)}{58})$$

Multiply 6i times 7+3i.

$$Re(\frac{6i\times 7+6\times 3i^{2}}{58})$$

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

$$Re(\frac{6i\times 7+6\times 3\left(-1\right)}{58})$$

Do the multiplications in 6i\times 7+6\times 3\left(-1\right). Reorder the terms.

$$Re(\frac{-18+42i}{58})$$

Divide -18+42i by 58 to get -\frac{9}{29}+\frac{21}{29}i.

$$Re(-\frac{9}{29}+\frac{21}{29}i)$$

The real part of -\frac{9}{29}+\frac{21}{29}i is -\frac{9}{29}.

$$-\frac{9}{29}$$