Solve (2+5i)(3+4i)=

$$( 2 + 5 i ) ( 3 + 4 i ) =$$

Evaluate

$$-14+23i$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$-14+23i$$

Real Part

$$-14$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(-14+23i)$$

The real part of -14+23i is -14.

$$-14$$