Solve (2+4i)(3-5i)+7(4-3i)

$$( 2 + 4 i ) ( 3 - 5 i ) + 7 ( 4 - 3 i )$$

Evaluate

$$54-19i$$

Solution Steps

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

$$2\times 3+2\times \left(-5i\right)+4i\times 3+4\left(-5\right)i^{2}+7\left(4-3i\right)$$

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

$$2\times 3+2\times \left(-5i\right)+4i\times 3+4\left(-5\right)\left(-1\right)+7\left(4-3i\right)$$

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

$$6-10i+12i+20+7\left(4-3i\right)$$

Combine the real and imaginary parts in 6-10i+12i+20.

$$6+20+\left(-10+12\right)i+7\left(4-3i\right)$$

Do the additions in 6+20+\left(-10+12\right)i.

$$26+2i+7\left(4-3i\right)$$

Multiply 7 times 4-3i.

$$26+2i+7\times 4+7\times \left(-3i\right)$$

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

$$26+2i+\left(28-21i\right)$$

Combine the real and imaginary parts.

$$26+28+\left(2-21\right)i$$

Do the additions.

$$54-19i$$

Real Part

$$54$$

Solution Steps

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

$$Re(2\times 3+2\times \left(-5i\right)+4i\times 3+4\left(-5\right)i^{2}+7\left(4-3i\right))$$

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

$$Re(2\times 3+2\times \left(-5i\right)+4i\times 3+4\left(-5\right)\left(-1\right)+7\left(4-3i\right))$$

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

$$Re(6-10i+12i+20+7\left(4-3i\right))$$

Combine the real and imaginary parts in 6-10i+12i+20.

$$Re(6+20+\left(-10+12\right)i+7\left(4-3i\right))$$

Do the additions in 6+20+\left(-10+12\right)i.

$$Re(26+2i+7\left(4-3i\right))$$

Multiply 7 times 4-3i.

$$Re(26+2i+7\times 4+7\times \left(-3i\right))$$

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

$$Re(26+2i+\left(28-21i\right))$$

Combine the real and imaginary parts in 26+2i+28-21i.

$$Re(26+28+\left(2-21\right)i)$$

Do the additions in 26+28+\left(2-21\right)i.

$$Re(54-19i)$$

The real part of 54-19i is 54.

$$54$$