Solve (1-2i)*(-5+4i)+(-2-2i)

$$( 1 - 2 i ) \cdot ( - 5 + 4 i ) + ( - 2 - 2 i )$$

Evaluate

$$1+12i$$

Solution Steps

Multiply complex numbers 1-2i and -5+4i like you multiply binomials.

$$1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4i^{2}+\left(-2-2i\right)$$

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

$$1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4\left(-1\right)+\left(-2-2i\right)$$

Do the multiplications in 1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4\left(-1\right).

$$-5+4i+10i+8+\left(-2-2i\right)$$

Combine the real and imaginary parts in -5+4i+10i+8.

$$-5+8+\left(4+10\right)i+\left(-2-2i\right)$$

Do the additions in -5+8+\left(4+10\right)i.

$$3+14i+\left(-2-2i\right)$$

Combine the real and imaginary parts.

$$3-2+\left(14-2\right)i$$

Do the additions.

$$1+12i$$

Real Part

$$1$$

Solution Steps

Multiply complex numbers 1-2i and -5+4i like you multiply binomials.

$$Re(1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4i^{2}+\left(-2-2i\right))$$

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

$$Re(1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4\left(-1\right)+\left(-2-2i\right))$$

Do the multiplications in 1\left(-5\right)+1\times \left(4i\right)-2i\left(-5\right)-2\times 4\left(-1\right).

$$Re(-5+4i+10i+8+\left(-2-2i\right))$$

Combine the real and imaginary parts in -5+4i+10i+8.

$$Re(-5+8+\left(4+10\right)i+\left(-2-2i\right))$$

Do the additions in -5+8+\left(4+10\right)i.

$$Re(3+14i+\left(-2-2i\right))$$

Combine the real and imaginary parts in 3+14i+\left(-2-2i\right).

$$Re(3-2+\left(14-2\right)i)$$

Do the additions in 3-2+\left(14-2\right)i.

$$Re(1+12i)$$

The real part of 1+12i is 1.

$$1$$