Solve (3-i)(2-i)-2i-sqrt{-9}

$$( 3 - i ) ( 2 - i ) - 2 i - \sqrt { - 9 }$$

Evaluate

$$5-10i$$

Solution Steps

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

$$3\times 2+3\left(-i\right)-i\times 2-\left(-i^{2}\right)-2i-\sqrt{-9}$$

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

$$3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right)-2i-\sqrt{-9}$$

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

$$6-3i-2i-1-2i-\sqrt{-9}$$

Combine the real and imaginary parts in 6-3i-2i-1.

$$6-1+\left(-3-2\right)i-2i-\sqrt{-9}$$

Do the additions in 6-1+\left(-3-2\right)i.

$$5-5i-2i-\sqrt{-9}$$

Combine the real and imaginary parts.

$$-\sqrt{-9}+5+\left(-5-2\right)i$$

Add -5 to -2.

$$-\sqrt{-9}+5-7i$$

Calculate the square root of -9 and get 3i.

$$-3i+5-7i$$

Combine the real and imaginary parts.

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

Add -3 to -7.

$$5-10i$$

Real Part

$$5$$

Solution Steps

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

$$Re(3\times 2+3\left(-i\right)-i\times 2-\left(-i^{2}\right)-2i-\sqrt{-9})$$

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

$$Re(3\times 2+3\left(-i\right)-i\times 2-\left(-\left(-1\right)\right)-2i-\sqrt{-9})$$

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

$$Re(6-3i-2i-1-2i-\sqrt{-9})$$

Combine the real and imaginary parts in 6-3i-2i-1.

$$Re(6-1+\left(-3-2\right)i-2i-\sqrt{-9})$$

Do the additions in 6-1+\left(-3-2\right)i.

$$Re(5-5i-2i-\sqrt{-9})$$

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

$$Re(-\sqrt{-9}+5+\left(-5-2\right)i)$$

Add -5 to -2.

$$Re(-\sqrt{-9}+5-7i)$$

Calculate the square root of -9 and get 3i.

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

Combine the real and imaginary parts in -3i+5-7i.

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

Add -3 to -7.

$$Re(5-10i)$$

The real part of 5-10i is 5.

$$5$$