Solve (9+2i)(3-7i)

$$( 9 + 2 i ) ( 3 - 7 i )$$

Evaluate

$$41-57i$$

Solution Steps

Multiply complex numbers 9+2i and 3-7i like you multiply binomials.

$$9\times 3+9\times \left(-7i\right)+2i\times 3+2\left(-7\right)i^{2}$$

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

$$9\times 3+9\times \left(-7i\right)+2i\times 3+2\left(-7\right)\left(-1\right)$$

Do the multiplications.

$$27-63i+6i+14$$

Combine the real and imaginary parts.

$$27+14+\left(-63+6\right)i$$

Do the additions.

$$41-57i$$

Real Part

$$41$$

Solution Steps

Multiply complex numbers 9+2i and 3-7i like you multiply binomials.

$$Re(9\times 3+9\times \left(-7i\right)+2i\times 3+2\left(-7\right)i^{2})$$

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

$$Re(9\times 3+9\times \left(-7i\right)+2i\times 3+2\left(-7\right)\left(-1\right))$$

Do the multiplications in 9\times 3+9\times \left(-7i\right)+2i\times 3+2\left(-7\right)\left(-1\right).

$$Re(27-63i+6i+14)$$

Combine the real and imaginary parts in 27-63i+6i+14.

$$Re(27+14+\left(-63+6\right)i)$$

Do the additions in 27+14+\left(-63+6\right)i.

$$Re(41-57i)$$

The real part of 41-57i is 41.

$$41$$