Solve 4+4i/-4+3i

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

Evaluate

$$-\frac{4}{25}-\frac{28}{25}i=-0.16-1.12i$$

Solution Steps

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

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

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

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

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

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

Multiply complex numbers 4+4i and -4-3i like you multiply binomials.

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

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

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

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

$$\frac{-16-12i-16i+12}{25}$$

Combine the real and imaginary parts in -16-12i-16i+12.

$$\frac{-16+12+\left(-12-16\right)i}{25}$$

Do the additions in -16+12+\left(-12-16\right)i.

$$\frac{-4-28i}{25}$$

Divide -4-28i by 25 to get -\frac{4}{25}-\frac{28}{25}i.

$$-\frac{4}{25}-\frac{28}{25}i$$

Real Part

$$-\frac{4}{25} = -0.16$$

Solution Steps

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

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

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

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

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

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

Multiply complex numbers 4+4i and -4-3i like you multiply binomials.

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

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

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

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

$$Re(\frac{-16-12i-16i+12}{25})$$

Combine the real and imaginary parts in -16-12i-16i+12.

$$Re(\frac{-16+12+\left(-12-16\right)i}{25})$$

Do the additions in -16+12+\left(-12-16\right)i.

$$Re(\frac{-4-28i}{25})$$

Divide -4-28i by 25 to get -\frac{4}{25}-\frac{28}{25}i.

$$Re(-\frac{4}{25}-\frac{28}{25}i)$$

The real part of -\frac{4}{25}-\frac{28}{25}i is -\frac{4}{25}.

$$-\frac{4}{25}$$