Solve (3-2i)(-4-7i)=

$$( 3 - 2 i ) ( - 4 - 7 i ) =$$

Evaluate

$$-26-13i$$

Solution Steps

Multiply complex numbers 3-2i and -4-7i like you multiply binomials.

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

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

$$3\left(-4\right)+3\times \left(-7i\right)-2i\left(-4\right)-2\left(-7\right)\left(-1\right)$$

Do the multiplications.

$$-12-21i+8i-14$$

Combine the real and imaginary parts.

$$-12-14+\left(-21+8\right)i$$

Do the additions.

$$-26-13i$$

Real Part

$$-26$$

Solution Steps

Multiply complex numbers 3-2i and -4-7i like you multiply binomials.

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

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

$$Re(3\left(-4\right)+3\times \left(-7i\right)-2i\left(-4\right)-2\left(-7\right)\left(-1\right))$$

Do the multiplications in 3\left(-4\right)+3\times \left(-7i\right)-2i\left(-4\right)-2\left(-7\right)\left(-1\right).

$$Re(-12-21i+8i-14)$$

Combine the real and imaginary parts in -12-21i+8i-14.

$$Re(-12-14+\left(-21+8\right)i)$$

Do the additions in -12-14+\left(-21+8\right)i.

$$Re(-26-13i)$$

The real part of -26-13i is -26.

$$-26$$