Solve (-2+5i)(-4+4i)

$$( - 2 + 5 i ) ( - 4 + 4 i )$$

Evaluate

$$-12-28i$$

Solution Steps

Multiply complex numbers -2+5i and -4+4i like you multiply binomials.

$$-2\left(-4\right)-2\times \left(4i\right)+5i\left(-4\right)+5\times 4i^{2}$$

By definition, i^{2} is -1.

$$-2\left(-4\right)-2\times \left(4i\right)+5i\left(-4\right)+5\times 4\left(-1\right)$$

Do the multiplications.

$$8-8i-20i-20$$

Combine the real and imaginary parts.

$$8-20+\left(-8-20\right)i$$

Do the additions.

$$-12-28i$$

Real Part

$$-12$$

Solution Steps

Multiply complex numbers -2+5i and -4+4i like you multiply binomials.

$$Re(-2\left(-4\right)-2\times \left(4i\right)+5i\left(-4\right)+5\times 4i^{2})$$

By definition, i^{2} is -1.

$$Re(-2\left(-4\right)-2\times \left(4i\right)+5i\left(-4\right)+5\times 4\left(-1\right))$$

Do the multiplications in -2\left(-4\right)-2\times \left(4i\right)+5i\left(-4\right)+5\times 4\left(-1\right).

$$Re(8-8i-20i-20)$$

Combine the real and imaginary parts in 8-8i-20i-20.

$$Re(8-20+\left(-8-20\right)i)$$

Do the additions in 8-20+\left(-8-20\right)i.

$$Re(-12-28i)$$

The real part of -12-28i is -12.

$$-12$$