Solve (6+7i)(3-7i)

$$( 6 + 7 i ) ( 3 - 7 i )$$

Evaluate

$$67-21i$$

Solution Steps

Multiply complex numbers 6+7i and 3-7i like you multiply binomials.

$$6\times 3+6\times \left(-7i\right)+7i\times 3+7\left(-7\right)i^{2}$$

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

$$6\times 3+6\times \left(-7i\right)+7i\times 3+7\left(-7\right)\left(-1\right)$$

Do the multiplications.

$$18-42i+21i+49$$

Combine the real and imaginary parts.

$$18+49+\left(-42+21\right)i$$

Do the additions.

$$67-21i$$

Real Part

$$67$$

Solution Steps

Multiply complex numbers 6+7i and 3-7i like you multiply binomials.

$$Re(6\times 3+6\times \left(-7i\right)+7i\times 3+7\left(-7\right)i^{2})$$

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

$$Re(6\times 3+6\times \left(-7i\right)+7i\times 3+7\left(-7\right)\left(-1\right))$$

Do the multiplications in 6\times 3+6\times \left(-7i\right)+7i\times 3+7\left(-7\right)\left(-1\right).

$$Re(18-42i+21i+49)$$

Combine the real and imaginary parts in 18-42i+21i+49.

$$Re(18+49+\left(-42+21\right)i)$$

Do the additions in 18+49+\left(-42+21\right)i.

$$Re(67-21i)$$

The real part of 67-21i is 67.

$$67$$