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

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

Evaluate

$$41-i$$

Solution Steps

Multiply complex numbers 2-5i and 3+7i like you multiply binomials.

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

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

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

Do the multiplications.

$$6+14i-15i+35$$

Combine the real and imaginary parts.

$$6+35+\left(14-15\right)i$$

Do the additions.

$$41-i$$

Real Part

$$41$$

Solution Steps

Multiply complex numbers 2-5i and 3+7i like you multiply binomials.

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

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

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

Do the multiplications in 2\times 3+2\times \left(7i\right)-5i\times 3-5\times 7\left(-1\right).

$$Re(6+14i-15i+35)$$

Combine the real and imaginary parts in 6+14i-15i+35.

$$Re(6+35+\left(14-15\right)i)$$

Do the additions in 6+35+\left(14-15\right)i.

$$Re(41-i)$$

The real part of 41-i is 41.

$$41$$