Solve (2i)^2*i^3*i^-4

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

Evaluate

$$4i$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 3 and -4 to get -1.

$$\left(2i\right)^{2}i^{-1}$$

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

$$-4i^{-1}$$

Calculate i to the power of -1 and get -i.

$$-4\left(-i\right)$$

Multiply -4 and -i to get 4i.

$$4i$$

Real Part

$$0$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 3 and -4 to get -1.

$$Re(\left(2i\right)^{2}i^{-1})$$

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

$$Re(-4i^{-1})$$

Calculate i to the power of -1 and get -i.

$$Re(-4\left(-i\right))$$

Multiply -4 and -i to get 4i.

$$Re(4i)$$

The real part of 4i is 0.

$$0$$