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

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

Evaluate

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

Solution Steps

Calculate 2+2i to the power of 3 and get -16+16i.

$$\frac{i\left(-16+16i\right)}{\left(2-4i\right)^{2}}$$

Multiply i and -16+16i to get -16-16i.

$$\frac{-16-16i}{\left(2-4i\right)^{2}}$$

Calculate 2-4i to the power of 2 and get -12-16i.

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

Multiply both numerator and denominator by the complex conjugate of the denominator, -12+16i.

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

Do the multiplications in \frac{\left(-16-16i\right)\left(-12+16i\right)}{\left(-12-16i\right)\left(-12+16i\right)}.

$$\frac{448-64i}{400}$$

Divide 448-64i by 400 to get \frac{28}{25}-\frac{4}{25}i.

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

Real Part

$$\frac{28}{25} = 1\frac{3}{25} = 1.12$$

Solution Steps

Calculate 2+2i to the power of 3 and get -16+16i.

$$Re(\frac{i\left(-16+16i\right)}{\left(2-4i\right)^{2}})$$

Multiply i and -16+16i to get -16-16i.

$$Re(\frac{-16-16i}{\left(2-4i\right)^{2}})$$

Calculate 2-4i to the power of 2 and get -12-16i.

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

Multiply both numerator and denominator of \frac{-16-16i}{-12-16i} by the complex conjugate of the denominator, -12+16i.

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

Do the multiplications in \frac{\left(-16-16i\right)\left(-12+16i\right)}{\left(-12-16i\right)\left(-12+16i\right)}.

$$Re(\frac{448-64i}{400})$$

Divide 448-64i by 400 to get \frac{28}{25}-\frac{4}{25}i.

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

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

$$\frac{28}{25}$$