Solve -3-3i/-3-i

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

Evaluate

$$\frac{6}{5}+\frac{3}{5}i=1.2+0.6i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -3+i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

$$\frac{\left(-3-3i\right)\left(-3+i\right)}{10}$$

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

$$\frac{-3\left(-3\right)-3i-3i\left(-3\right)-3i^{2}}{10}$$

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

$$\frac{-3\left(-3\right)-3i-3i\left(-3\right)-3\left(-1\right)}{10}$$

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

$$\frac{9-3i+9i+3}{10}$$

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

$$\frac{9+3+\left(-3+9\right)i}{10}$$

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

$$\frac{12+6i}{10}$$

Divide 12+6i by 10 to get \frac{6}{5}+\frac{3}{5}i.

$$\frac{6}{5}+\frac{3}{5}i$$

Real Part

$$\frac{6}{5} = 1\frac{1}{5} = 1.2$$

Solution Steps

Multiply both numerator and denominator of \frac{-3-3i}{-3-i} by the complex conjugate of the denominator, -3+i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

$$Re(\frac{\left(-3-3i\right)\left(-3+i\right)}{10})$$

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

$$Re(\frac{-3\left(-3\right)-3i-3i\left(-3\right)-3i^{2}}{10})$$

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

$$Re(\frac{-3\left(-3\right)-3i-3i\left(-3\right)-3\left(-1\right)}{10})$$

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

$$Re(\frac{9-3i+9i+3}{10})$$

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

$$Re(\frac{9+3+\left(-3+9\right)i}{10})$$

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

$$Re(\frac{12+6i}{10})$$

Divide 12+6i by 10 to get \frac{6}{5}+\frac{3}{5}i.

$$Re(\frac{6}{5}+\frac{3}{5}i)$$

The real part of \frac{6}{5}+\frac{3}{5}i is \frac{6}{5}.

$$\frac{6}{5}$$