Solve (6+7i)(1-2i)

$$( 6 + 7 i ) ( 1 - 2 i )$$

Evaluate

$$20-5i$$

Solution Steps

Multiply complex numbers 6+7i and 1-2i like you multiply binomials.

$$6\times 1+6\times \left(-2i\right)+7i\times 1+7\left(-2\right)i^{2}$$

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

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

Do the multiplications.

$$6-12i+7i+14$$

Combine the real and imaginary parts.

$$6+14+\left(-12+7\right)i$$

Do the additions.

$$20-5i$$

Real Part

$$20$$

Solution Steps

Multiply complex numbers 6+7i and 1-2i like you multiply binomials.

$$Re(6\times 1+6\times \left(-2i\right)+7i\times 1+7\left(-2\right)i^{2})$$

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

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

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

$$Re(6-12i+7i+14)$$

Combine the real and imaginary parts in 6-12i+7i+14.

$$Re(6+14+\left(-12+7\right)i)$$

Do the additions in 6+14+\left(-12+7\right)i.

$$Re(20-5i)$$

The real part of 20-5i is 20.

$$20$$