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

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

Evaluate

$$\frac{224}{25}-\frac{32}{25}i=8.96-1.28i$$

Solution Steps

Calculate 4+4i to the power of 3 and get -128+128i.

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

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

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

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

$$\frac{-128-128i}{-12-16i}$$

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

$$\frac{\left(-128-128i\right)\left(-12+16i\right)}{\left(-12-16i\right)\left(-12+16i\right)}$$

Do the multiplications in \frac{\left(-128-128i\right)\left(-12+16i\right)}{\left(-12-16i\right)\left(-12+16i\right)}.

$$\frac{3584-512i}{400}$$

Divide 3584-512i by 400 to get \frac{224}{25}-\frac{32}{25}i.

$$\frac{224}{25}-\frac{32}{25}i$$

Real Part

$$\frac{224}{25} = 8\frac{24}{25} = 8.96$$

Solution Steps

Calculate 4+4i to the power of 3 and get -128+128i.

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

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

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

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

$$Re(\frac{-128-128i}{-12-16i})$$

Multiply both numerator and denominator of \frac{-128-128i}{-12-16i} by the complex conjugate of the denominator, -12+16i.

$$Re(\frac{\left(-128-128i\right)\left(-12+16i\right)}{\left(-12-16i\right)\left(-12+16i\right)})$$

Do the multiplications in \frac{\left(-128-128i\right)\left(-12+16i\right)}{\left(-12-16i\right)\left(-12+16i\right)}.

$$Re(\frac{3584-512i}{400})$$

Divide 3584-512i by 400 to get \frac{224}{25}-\frac{32}{25}i.

$$Re(\frac{224}{25}-\frac{32}{25}i)$$

The real part of \frac{224}{25}-\frac{32}{25}i is \frac{224}{25}.

$$\frac{224}{25}$$