Solve (2-5i)(1-i)-(3-i)(3+i)

$$( 2 - 5 i ) ( 1 - i ) - ( 3 - i ) ( 3 + i )$$

Evaluate

$$-13-7i$$

Solution Steps

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

$$2\times 1+2\left(-i\right)-5i-5\left(-1\right)i^{2}-\left(3-i\right)\left(3+i\right)$$

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

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

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

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

Combine the real and imaginary parts in 2-2i-5i-5.

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

Do the additions in 2-5+\left(-2-5\right)i.

$$-3-7i-\left(3-i\right)\left(3+i\right)$$

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

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

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

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

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

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

Combine the real and imaginary parts in 9+3i-3i+1.

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

Do the additions in 9+1+\left(3-3\right)i.

$$-3-7i-10$$

Subtract 10 from -3-7i by subtracting corresponding real and imaginary parts.

$$-3-10-7i$$

Subtract 10 from -3 to get -13.

$$-13-7i$$

Real Part

$$-13$$

Solution Steps

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

$$Re(2\times 1+2\left(-i\right)-5i-5\left(-1\right)i^{2}-\left(3-i\right)\left(3+i\right))$$

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

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

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

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

Combine the real and imaginary parts in 2-2i-5i-5.

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

Do the additions in 2-5+\left(-2-5\right)i.

$$Re(-3-7i-\left(3-i\right)\left(3+i\right))$$

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

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

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

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

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

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

Combine the real and imaginary parts in 9+3i-3i+1.

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

Do the additions in 9+1+\left(3-3\right)i.

$$Re(-3-7i-10)$$

Subtract 10 from -3-7i by subtracting corresponding real and imaginary parts.

$$Re(-3-10-7i)$$

Subtract 10 from -3 to get -13.

$$Re(-13-7i)$$

The real part of -13-7i is -13.

$$-13$$