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

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

Evaluate

$$34-13i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$6+8i-21i+28$$

Combine the real and imaginary parts.

$$6+28+\left(8-21\right)i$$

Do the additions.

$$34-13i$$

Real Part

$$34$$

Solution Steps

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

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

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

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

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

$$Re(6+8i-21i+28)$$

Combine the real and imaginary parts in 6+8i-21i+28.

$$Re(6+28+\left(8-21\right)i)$$

Do the additions in 6+28+\left(8-21\right)i.

$$Re(34-13i)$$

The real part of 34-13i is 34.

$$34$$