Solve (4-2i)-(6+i^3)+i^2

$$( 4 - 2 i ) - ( 6 + i ^ { 3 } ) + i ^ { 2 }$$

Evaluate

$$-3-i$$

Solution Steps

Calculate i to the power of 3 and get -i.

$$4-2i-\left(6-i\right)+i^{2}$$

Subtract 6-i from 4-2i to get -2-i.

$$-2-i+i^{2}$$

Calculate i to the power of 2 and get -1.

$$-2-i-1$$

Subtract 1 from -2-i to get -3-i.

$$-3-i$$

Real Part

$$-3$$

Solution Steps

Calculate i to the power of 3 and get -i.

$$Re(4-2i-\left(6-i\right)+i^{2})$$

Subtract 6-i from 4-2i to get -2-i.

$$Re(-2-i+i^{2})$$

Calculate i to the power of 2 and get -1.

$$Re(-2-i-1)$$

Subtract 1 from -2-i to get -3-i.

$$Re(-3-i)$$

The real part of -3-i is -3.

$$-3$$