Solve (-1+5i)(4+i)

$$( - 1 + 5 i ) ( 4 + i )$$

Evaluate

$$-9+19i$$

Solution Steps

Multiply complex numbers -1+5i and 4+i like you multiply binomials.

$$-4-i+5i\times 4+5i^{2}$$

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

$$-4-i+5i\times 4+5\left(-1\right)$$

Do the multiplications.

$$-4-i+20i-5$$

Combine the real and imaginary parts.

$$-4-5+\left(-1+20\right)i$$

Do the additions.

$$-9+19i$$

Real Part

$$-9$$

Solution Steps

Multiply complex numbers -1+5i and 4+i like you multiply binomials.

$$Re(-4-i+5i\times 4+5i^{2})$$

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

$$Re(-4-i+5i\times 4+5\left(-1\right))$$

Do the multiplications in -4-i+5i\times 4+5\left(-1\right).

$$Re(-4-i+20i-5)$$

Combine the real and imaginary parts in -4-i+20i-5.

$$Re(-4-5+\left(-1+20\right)i)$$

Do the additions in -4-5+\left(-1+20\right)i.

$$Re(-9+19i)$$

The real part of -9+19i is -9.

$$-9$$