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

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

Evaluate

$$-10-30i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$-20-10i-20i+10$$

Combine the real and imaginary parts.

$$-20+10+\left(-10-20\right)i$$

Do the additions.

$$-10-30i$$

Real Part

$$-10$$

Solution Steps

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

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

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

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

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

$$Re(-20-10i-20i+10)$$

Combine the real and imaginary parts in -20-10i-20i+10.

$$Re(-20+10+\left(-10-20\right)i)$$

Do the additions in -20+10+\left(-10-20\right)i.

$$Re(-10-30i)$$

The real part of -10-30i is -10.

$$-10$$