Solve (-4+4i4^-2)*4^2

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

Evaluate

$$-64+4i$$

Solution Steps

Calculate 4 to the power of -2 and get \frac{1}{16}.

$$\left(-4+4i\times \frac{1}{16}\right)\times 4^{2}$$

Multiply 4i and \frac{1}{16} to get \frac{1}{4}i.

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

Calculate 4 to the power of 2 and get 16.

$$\left(-4+\frac{1}{4}i\right)\times 16$$

Multiply -4+\frac{1}{4}i and 16 to get -64+4i.

$$-64+4i$$

Real Part

$$-64$$

Solution Steps

Calculate 4 to the power of -2 and get \frac{1}{16}.

$$Re(\left(-4+4i\times \frac{1}{16}\right)\times 4^{2})$$

Multiply 4i and \frac{1}{16} to get \frac{1}{4}i.

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

Calculate 4 to the power of 2 and get 16.

$$Re(\left(-4+\frac{1}{4}i\right)\times 16)$$

Multiply -4+\frac{1}{4}i and 16 to get -64+4i.

$$Re(-64+4i)$$

The real part of -64+4i is -64.

$$-64$$