Solve (4-4i)(3i)/-1+3i

$$\frac { ( 4 - 4 i ) ( 3 i ) } { - 1 + 3 i }$$

Evaluate

$$\frac{12}{5}-\frac{24}{5}i=2.4-4.8i$$

Solution Steps

Multiply 4-4i times 3i.

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

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

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

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

$$\frac{12+12i}{-1+3i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, -1-3i.

$$\frac{\left(12+12i\right)\left(-1-3i\right)}{\left(-1+3i\right)\left(-1-3i\right)}$$

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

$$\frac{\left(12+12i\right)\left(-1-3i\right)}{\left(-1\right)^{2}-3^{2}i^{2}}$$

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

$$\frac{\left(12+12i\right)\left(-1-3i\right)}{10}$$

Multiply complex numbers 12+12i and -1-3i like you multiply binomials.

$$\frac{12\left(-1\right)+12\times \left(-3i\right)+12i\left(-1\right)+12\left(-3\right)i^{2}}{10}$$

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

$$\frac{12\left(-1\right)+12\times \left(-3i\right)+12i\left(-1\right)+12\left(-3\right)\left(-1\right)}{10}$$

Do the multiplications in 12\left(-1\right)+12\times \left(-3i\right)+12i\left(-1\right)+12\left(-3\right)\left(-1\right).

$$\frac{-12-36i-12i+36}{10}$$

Combine the real and imaginary parts in -12-36i-12i+36.

$$\frac{-12+36+\left(-36-12\right)i}{10}$$

Do the additions in -12+36+\left(-36-12\right)i.

$$\frac{24-48i}{10}$$

Divide 24-48i by 10 to get \frac{12}{5}-\frac{24}{5}i.

$$\frac{12}{5}-\frac{24}{5}i$$

Real Part

$$\frac{12}{5} = 2\frac{2}{5} = 2.4$$

Solution Steps

Multiply 4-4i times 3i.

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

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

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

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

$$Re(\frac{12+12i}{-1+3i})$$

Multiply both numerator and denominator of \frac{12+12i}{-1+3i} by the complex conjugate of the denominator, -1-3i.

$$Re(\frac{\left(12+12i\right)\left(-1-3i\right)}{\left(-1+3i\right)\left(-1-3i\right)})$$

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

$$Re(\frac{\left(12+12i\right)\left(-1-3i\right)}{\left(-1\right)^{2}-3^{2}i^{2}})$$

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

$$Re(\frac{\left(12+12i\right)\left(-1-3i\right)}{10})$$

Multiply complex numbers 12+12i and -1-3i like you multiply binomials.

$$Re(\frac{12\left(-1\right)+12\times \left(-3i\right)+12i\left(-1\right)+12\left(-3\right)i^{2}}{10})$$

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

$$Re(\frac{12\left(-1\right)+12\times \left(-3i\right)+12i\left(-1\right)+12\left(-3\right)\left(-1\right)}{10})$$

Do the multiplications in 12\left(-1\right)+12\times \left(-3i\right)+12i\left(-1\right)+12\left(-3\right)\left(-1\right).

$$Re(\frac{-12-36i-12i+36}{10})$$

Combine the real and imaginary parts in -12-36i-12i+36.

$$Re(\frac{-12+36+\left(-36-12\right)i}{10})$$

Do the additions in -12+36+\left(-36-12\right)i.

$$Re(\frac{24-48i}{10})$$

Divide 24-48i by 10 to get \frac{12}{5}-\frac{24}{5}i.

$$Re(\frac{12}{5}-\frac{24}{5}i)$$

The real part of \frac{12}{5}-\frac{24}{5}i is \frac{12}{5}.

$$\frac{12}{5}$$