Solve 5+5i/-3-4i

$$\frac{ 5+5i }{ -3-4i }$$

Evaluate

$$-\frac{7}{5}+\frac{1}{5}i=-1.4+0.2i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -3+4i.

$$\frac{\left(5+5i\right)\left(-3+4i\right)}{\left(-3-4i\right)\left(-3+4i\right)}$$

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

$$\frac{\left(5+5i\right)\left(-3+4i\right)}{\left(-3\right)^{2}-4^{2}i^{2}}$$

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

$$\frac{\left(5+5i\right)\left(-3+4i\right)}{25}$$

Multiply complex numbers 5+5i and -3+4i like you multiply binomials.

$$\frac{5\left(-3\right)+5\times \left(4i\right)+5i\left(-3\right)+5\times 4i^{2}}{25}$$

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

$$\frac{5\left(-3\right)+5\times \left(4i\right)+5i\left(-3\right)+5\times 4\left(-1\right)}{25}$$

Do the multiplications in 5\left(-3\right)+5\times \left(4i\right)+5i\left(-3\right)+5\times 4\left(-1\right).

$$\frac{-15+20i-15i-20}{25}$$

Combine the real and imaginary parts in -15+20i-15i-20.

$$\frac{-15-20+\left(20-15\right)i}{25}$$

Do the additions in -15-20+\left(20-15\right)i.

$$\frac{-35+5i}{25}$$

Divide -35+5i by 25 to get -\frac{7}{5}+\frac{1}{5}i.

$$-\frac{7}{5}+\frac{1}{5}i$$

Real Part

$$-\frac{7}{5} = -1\frac{2}{5} = -1.4$$

Solution Steps

Multiply both numerator and denominator of \frac{5+5i}{-3-4i} by the complex conjugate of the denominator, -3+4i.

$$Re(\frac{\left(5+5i\right)\left(-3+4i\right)}{\left(-3-4i\right)\left(-3+4i\right)})$$

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

$$Re(\frac{\left(5+5i\right)\left(-3+4i\right)}{\left(-3\right)^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{\left(5+5i\right)\left(-3+4i\right)}{25})$$

Multiply complex numbers 5+5i and -3+4i like you multiply binomials.

$$Re(\frac{5\left(-3\right)+5\times \left(4i\right)+5i\left(-3\right)+5\times 4i^{2}}{25})$$

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

$$Re(\frac{5\left(-3\right)+5\times \left(4i\right)+5i\left(-3\right)+5\times 4\left(-1\right)}{25})$$

Do the multiplications in 5\left(-3\right)+5\times \left(4i\right)+5i\left(-3\right)+5\times 4\left(-1\right).

$$Re(\frac{-15+20i-15i-20}{25})$$

Combine the real and imaginary parts in -15+20i-15i-20.

$$Re(\frac{-15-20+\left(20-15\right)i}{25})$$

Do the additions in -15-20+\left(20-15\right)i.

$$Re(\frac{-35+5i}{25})$$

Divide -35+5i by 25 to get -\frac{7}{5}+\frac{1}{5}i.

$$Re(-\frac{7}{5}+\frac{1}{5}i)$$

The real part of -\frac{7}{5}+\frac{1}{5}i is -\frac{7}{5}.

$$-\frac{7}{5}$$