Solve (5+7i)(6-2i)

$$( 5 + 7 i ) ( 6 - 2 i )$$

Evaluate

$$44+32i$$

Solution Steps

Multiply complex numbers 5+7i and 6-2i like you multiply binomials.

$$5\times 6+5\times \left(-2i\right)+7i\times 6+7\left(-2\right)i^{2}$$

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

$$5\times 6+5\times \left(-2i\right)+7i\times 6+7\left(-2\right)\left(-1\right)$$

Do the multiplications.

$$30-10i+42i+14$$

Combine the real and imaginary parts.

$$30+14+\left(-10+42\right)i$$

Do the additions.

$$44+32i$$

Real Part

$$44$$

Solution Steps

Multiply complex numbers 5+7i and 6-2i like you multiply binomials.

$$Re(5\times 6+5\times \left(-2i\right)+7i\times 6+7\left(-2\right)i^{2})$$

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

$$Re(5\times 6+5\times \left(-2i\right)+7i\times 6+7\left(-2\right)\left(-1\right))$$

Do the multiplications in 5\times 6+5\times \left(-2i\right)+7i\times 6+7\left(-2\right)\left(-1\right).

$$Re(30-10i+42i+14)$$

Combine the real and imaginary parts in 30-10i+42i+14.

$$Re(30+14+\left(-10+42\right)i)$$

Do the additions in 30+14+\left(-10+42\right)i.

$$Re(44+32i)$$

The real part of 44+32i is 44.

$$44$$