Solve 2+2i/-3-3i=

$$\frac { 2 + 2 i } { - 3 - 3 i } =$$

Evaluate

$$-\frac{2}{3}\approx -0.666666667$$

Solution Steps

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

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

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

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

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

$$\frac{\left(2+2i\right)\left(-3+3i\right)}{18}$$

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

$$\frac{2\left(-3\right)+2\times \left(3i\right)+2i\left(-3\right)+2\times 3i^{2}}{18}$$

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

$$\frac{2\left(-3\right)+2\times \left(3i\right)+2i\left(-3\right)+2\times 3\left(-1\right)}{18}$$

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

$$\frac{-6+6i-6i-6}{18}$$

Combine the real and imaginary parts in -6+6i-6i-6.

$$\frac{-6-6+\left(6-6\right)i}{18}$$

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

$$\frac{-12}{18}$$

Divide -12 by 18 to get -\frac{2}{3}.

$$-\frac{2}{3}$$

Real Part

$$-\frac{2}{3} = -0.6666666666666666$$

Solution Steps

Multiply both numerator and denominator of \frac{2+2i}{-3-3i} by the complex conjugate of the denominator, -3+3i.

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

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

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

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

$$Re(\frac{\left(2+2i\right)\left(-3+3i\right)}{18})$$

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

$$Re(\frac{2\left(-3\right)+2\times \left(3i\right)+2i\left(-3\right)+2\times 3i^{2}}{18})$$

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

$$Re(\frac{2\left(-3\right)+2\times \left(3i\right)+2i\left(-3\right)+2\times 3\left(-1\right)}{18})$$

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

$$Re(\frac{-6+6i-6i-6}{18})$$

Combine the real and imaginary parts in -6+6i-6i-6.

$$Re(\frac{-6-6+\left(6-6\right)i}{18})$$

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

$$Re(\frac{-12}{18})$$

Divide -12 by 18 to get -\frac{2}{3}.

$$Re(-\frac{2}{3})$$

The real part of -\frac{2}{3} is -\frac{2}{3}.

$$-\frac{2}{3}$$