Solve (32-32i)(2+2i)/3(-4+4i)^2

$$\frac { ( 32 - 32 i ) ( 2 + 2 i ) } { 3 ( - 4 + 4 i ) ^ { 2 } }$$

Evaluate

$$\frac{4}{3}i\approx 1.333333333i$$

Solution Steps

Multiply 32-32i and 2+2i to get 128.

$$\frac{128}{3\left(-4+4i\right)^{2}}$$

Calculate -4+4i to the power of 2 and get -32i.

$$\frac{128}{3\times \left(-32i\right)}$$

Multiply 3 and -32i to get -96i.

$$\frac{128}{-96i}$$

Multiply both numerator and denominator by imaginary unit i.

$$\frac{128i}{96}$$

Divide 128i by 96 to get \frac{4}{3}i.

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

Real Part

$$0$$

Solution Steps

Multiply 32-32i and 2+2i to get 128.

$$Re(\frac{128}{3\left(-4+4i\right)^{2}})$$

Calculate -4+4i to the power of 2 and get -32i.

$$Re(\frac{128}{3\times \left(-32i\right)})$$

Multiply 3 and -32i to get -96i.

$$Re(\frac{128}{-96i})$$

Multiply both numerator and denominator of \frac{128}{-96i} by imaginary unit i.

$$Re(\frac{128i}{96})$$

Divide 128i by 96 to get \frac{4}{3}i.

$$Re(\frac{4}{3}i)$$

The real part of \frac{4}{3}i is 0.

$$0$$