Solve -12i/14+8i

$$\frac { - 12 i } { 14 + 8 i }$$

Evaluate

$$-\frac{24}{65}-\frac{42}{65}i\approx -0.369230769-0.646153846i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 14-8i.

$$\frac{-12i\left(14-8i\right)}{\left(14+8i\right)\left(14-8i\right)}$$

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

$$\frac{-12i\left(14-8i\right)}{14^{2}-8^{2}i^{2}}$$

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

$$\frac{-12i\left(14-8i\right)}{260}$$

Multiply -12i times 14-8i.

$$\frac{-12i\times 14-12\left(-8\right)i^{2}}{260}$$

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

$$\frac{-12i\times 14-12\left(-8\right)\left(-1\right)}{260}$$

Do the multiplications in -12i\times 14-12\left(-8\right)\left(-1\right). Reorder the terms.

$$\frac{-96-168i}{260}$$

Divide -96-168i by 260 to get -\frac{24}{65}-\frac{42}{65}i.

$$-\frac{24}{65}-\frac{42}{65}i$$

Real Part

$$-\frac{24}{65} = -0.36923076923076925$$

Solution Steps

Multiply both numerator and denominator of \frac{-12i}{14+8i} by the complex conjugate of the denominator, 14-8i.

$$Re(\frac{-12i\left(14-8i\right)}{\left(14+8i\right)\left(14-8i\right)})$$

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

$$Re(\frac{-12i\left(14-8i\right)}{14^{2}-8^{2}i^{2}})$$

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

$$Re(\frac{-12i\left(14-8i\right)}{260})$$

Multiply -12i times 14-8i.

$$Re(\frac{-12i\times 14-12\left(-8\right)i^{2}}{260})$$

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

$$Re(\frac{-12i\times 14-12\left(-8\right)\left(-1\right)}{260})$$

Do the multiplications in -12i\times 14-12\left(-8\right)\left(-1\right). Reorder the terms.

$$Re(\frac{-96-168i}{260})$$

Divide -96-168i by 260 to get -\frac{24}{65}-\frac{42}{65}i.

$$Re(-\frac{24}{65}-\frac{42}{65}i)$$

The real part of -\frac{24}{65}-\frac{42}{65}i is -\frac{24}{65}.

$$-\frac{24}{65}$$