Solve (-4-7i)(3-5i)

$$( - 4 - 7 i ) ( 3 - 5 i )$$

Evaluate

$$-47-i$$

Solution Steps

Multiply complex numbers -4-7i and 3-5i like you multiply binomials.

$$-4\times 3-4\times \left(-5i\right)-7i\times 3-7\left(-5\right)i^{2}$$

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

$$-4\times 3-4\times \left(-5i\right)-7i\times 3-7\left(-5\right)\left(-1\right)$$

Do the multiplications.

$$-12+20i-21i-35$$

Combine the real and imaginary parts.

$$-12-35+\left(20-21\right)i$$

Do the additions.

$$-47-i$$

Real Part

$$-47$$

Solution Steps

Multiply complex numbers -4-7i and 3-5i like you multiply binomials.

$$Re(-4\times 3-4\times \left(-5i\right)-7i\times 3-7\left(-5\right)i^{2})$$

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

$$Re(-4\times 3-4\times \left(-5i\right)-7i\times 3-7\left(-5\right)\left(-1\right))$$

Do the multiplications in -4\times 3-4\times \left(-5i\right)-7i\times 3-7\left(-5\right)\left(-1\right).

$$Re(-12+20i-21i-35)$$

Combine the real and imaginary parts in -12+20i-21i-35.

$$Re(-12-35+\left(20-21\right)i)$$

Do the additions in -12-35+\left(20-21\right)i.

$$Re(-47-i)$$

The real part of -47-i is -47.

$$-47$$