Solve (2-5i)(1+3i)+i(4-i)

$$( 2 - 5 i ) ( 1 + 3 i ) + i ( 4 - i )$$

Evaluate

$$18+5i$$

Solution Steps

Multiply complex numbers 2-5i and 1+3i like you multiply binomials.

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

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

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

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

$$2+6i-5i+15+i\left(4-i\right)$$

Combine the real and imaginary parts in 2+6i-5i+15.

$$2+15+\left(6-5\right)i+i\left(4-i\right)$$

Do the additions in 2+15+\left(6-5\right)i.

$$17+i+i\left(4-i\right)$$

Multiply i times 4-i.

$$17+i+4i-i^{2}$$

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

$$17+i+4i-\left(-1\right)$$

Do the multiplications in 4i-\left(-1\right). Reorder the terms.

$$17+i+\left(1+4i\right)$$

Combine the real and imaginary parts.

$$17+1+\left(1+4\right)i$$

Do the additions.

$$18+5i$$

Real Part

$$18$$

Solution Steps

Multiply complex numbers 2-5i and 1+3i like you multiply binomials.

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

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

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

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

$$Re(2+6i-5i+15+i\left(4-i\right))$$

Combine the real and imaginary parts in 2+6i-5i+15.

$$Re(2+15+\left(6-5\right)i+i\left(4-i\right))$$

Do the additions in 2+15+\left(6-5\right)i.

$$Re(17+i+i\left(4-i\right))$$

Multiply i times 4-i.

$$Re(17+i+4i-i^{2})$$

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

$$Re(17+i+4i-\left(-1\right))$$

Do the multiplications in 4i-\left(-1\right). Reorder the terms.

$$Re(17+i+\left(1+4i\right))$$

Combine the real and imaginary parts in 17+i+1+4i.

$$Re(17+1+\left(1+4\right)i)$$

Do the additions in 17+1+\left(1+4\right)i.

$$Re(18+5i)$$

The real part of 18+5i is 18.

$$18$$