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

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

Evaluate

$$-13-35i$$

Solution Steps

Multiply 5i times 7-i.

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

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

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

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

$$2-\left(5+35i\right)-\left(3-i\right)\left(3+i\right)$$

Subtract 5+35i from 2 by subtracting corresponding real and imaginary parts.

$$2-5+35i-\left(3-i\right)\left(3+i\right)$$

Subtract 5 from 2.

$$-3-35i-\left(3-i\right)\left(3+i\right)$$

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

$$-3-35i-\left(3\times 3+3i-i\times 3-i^{2}\right)$$

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

$$-3-35i-\left(3\times 3+3i-i\times 3-\left(-1\right)\right)$$

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

$$-3-35i-\left(9+3i-3i+1\right)$$

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

$$-3-35i-\left(9+1+\left(3-3\right)i\right)$$

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

$$-3-35i-10$$

Subtract 10 from -3-35i by subtracting corresponding real and imaginary parts.

$$-3-10-35i$$

Subtract 10 from -3 to get -13.

$$-13-35i$$

Real Part

$$-13$$

Solution Steps

Multiply 5i times 7-i.

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

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

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

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

$$Re(2-\left(5+35i\right)-\left(3-i\right)\left(3+i\right))$$

Subtract 5+35i from 2 by subtracting corresponding real and imaginary parts.

$$Re(2-5+35i-\left(3-i\right)\left(3+i\right))$$

Subtract 5 from 2.

$$Re(-3-35i-\left(3-i\right)\left(3+i\right))$$

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

$$Re(-3-35i-\left(3\times 3+3i-i\times 3-i^{2}\right))$$

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

$$Re(-3-35i-\left(3\times 3+3i-i\times 3-\left(-1\right)\right))$$

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

$$Re(-3-35i-\left(9+3i-3i+1\right))$$

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

$$Re(-3-35i-\left(9+1+\left(3-3\right)i\right))$$

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

$$Re(-3-35i-10)$$

Subtract 10 from -3-35i by subtracting corresponding real and imaginary parts.

$$Re(-3-10-35i)$$

Subtract 10 from -3 to get -13.

$$Re(-13-35i)$$

The real part of -13-35i is -13.

$$-13$$