Solve (-3-4i)*(2+5i)

$$( - 3 - 4 i ) \times ( 2 + 5 i )$$

Evaluate

$$14-23i$$

Solution Steps

Multiply complex numbers -3-4i and 2+5i like you multiply binomials.

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

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

$$-3\times 2-3\times \left(5i\right)-4i\times 2-4\times 5\left(-1\right)$$

Do the multiplications.

$$-6-15i-8i+20$$

Combine the real and imaginary parts.

$$-6+20+\left(-15-8\right)i$$

Do the additions.

$$14-23i$$

Real Part

$$14$$

Solution Steps

Multiply complex numbers -3-4i and 2+5i like you multiply binomials.

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

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

$$Re(-3\times 2-3\times \left(5i\right)-4i\times 2-4\times 5\left(-1\right))$$

Do the multiplications in -3\times 2-3\times \left(5i\right)-4i\times 2-4\times 5\left(-1\right).

$$Re(-6-15i-8i+20)$$

Combine the real and imaginary parts in -6-15i-8i+20.

$$Re(-6+20+\left(-15-8\right)i)$$

Do the additions in -6+20+\left(-15-8\right)i.

$$Re(14-23i)$$

The real part of 14-23i is 14.

$$14$$