Solve (2+3i)(-1-i)+3/i

$$( 2 + 3 i ) ( - 1 - i ) + \frac { 3 } { i }$$

Evaluate

$$1-8i$$

Solution Steps

Multiply complex numbers 2+3i and -1-i like you multiply binomials.

$$2\left(-1\right)+2\left(-i\right)+3i\left(-1\right)+3\left(-1\right)i^{2}+\frac{3}{i}$$

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

$$2\left(-1\right)+2\left(-i\right)+3i\left(-1\right)+3\left(-1\right)\left(-1\right)+\frac{3}{i}$$

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

$$-2-2i-3i+3+\frac{3}{i}$$

Combine the real and imaginary parts in -2-2i-3i+3.

$$-2+3+\left(-2-3\right)i+\frac{3}{i}$$

Do the additions in -2+3+\left(-2-3\right)i.

$$1-5i+\frac{3}{i}$$

Multiply both numerator and denominator of \frac{3}{i} by imaginary unit i.

$$1-5i+\frac{3i}{1i^{2}}$$

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

$$1-5i+\frac{3i}{-1}$$

Divide 3i by -1 to get -3i.

$$1-5i-3i$$

Combine the real and imaginary parts.

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

Add -5 to -3.

$$1-8i$$

Real Part

$$1$$

Solution Steps

Multiply complex numbers 2+3i and -1-i like you multiply binomials.

$$Re(2\left(-1\right)+2\left(-i\right)+3i\left(-1\right)+3\left(-1\right)i^{2}+\frac{3}{i})$$

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

$$Re(2\left(-1\right)+2\left(-i\right)+3i\left(-1\right)+3\left(-1\right)\left(-1\right)+\frac{3}{i})$$

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

$$Re(-2-2i-3i+3+\frac{3}{i})$$

Combine the real and imaginary parts in -2-2i-3i+3.

$$Re(-2+3+\left(-2-3\right)i+\frac{3}{i})$$

Do the additions in -2+3+\left(-2-3\right)i.

$$Re(1-5i+\frac{3}{i})$$

Multiply both numerator and denominator of \frac{3}{i} by imaginary unit i.

$$Re(1-5i+\frac{3i}{1i^{2}})$$

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

$$Re(1-5i+\frac{3i}{-1})$$

Divide 3i by -1 to get -3i.

$$Re(1-5i-3i)$$

Combine the real and imaginary parts in 1-5i-3i.

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

Add -5 to -3.

$$Re(1-8i)$$

The real part of 1-8i is 1.

$$1$$