Solve -3-3i/-3+i

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

Evaluate

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

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.

$$\frac{-3\left(-3\right)-3\left(-i\right)-3i\left(-3\right)-3\left(-1\right)i^{2}}{10}$$

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

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

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

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

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

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

Real Part

$$\frac{3}{5} = 0.6$$

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.

$$Re(\frac{-3\left(-3\right)-3\left(-i\right)-3i\left(-3\right)-3\left(-1\right)i^{2}}{10})$$

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

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

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

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

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

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

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

$$\frac{3}{5}$$