Solve -3+5i/3-2i

$$\frac { - 3 + 5 i } { 3 - 2 i }$$

Evaluate

$$-\frac{19}{13}+\frac{9}{13}i\approx -1.461538462+0.692307692i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+2i.

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

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

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

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

$$\frac{\left(-3+5i\right)\left(3+2i\right)}{13}$$

Multiply complex numbers -3+5i and 3+2i like you multiply binomials.

$$\frac{-3\times 3-3\times \left(2i\right)+5i\times 3+5\times 2i^{2}}{13}$$

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

$$\frac{-3\times 3-3\times \left(2i\right)+5i\times 3+5\times 2\left(-1\right)}{13}$$

Do the multiplications in -3\times 3-3\times \left(2i\right)+5i\times 3+5\times 2\left(-1\right).

$$\frac{-9-6i+15i-10}{13}$$

Combine the real and imaginary parts in -9-6i+15i-10.

$$\frac{-9-10+\left(-6+15\right)i}{13}$$

Do the additions in -9-10+\left(-6+15\right)i.

$$\frac{-19+9i}{13}$$

Divide -19+9i by 13 to get -\frac{19}{13}+\frac{9}{13}i.

$$-\frac{19}{13}+\frac{9}{13}i$$

Real Part

$$-\frac{19}{13} = -1\frac{6}{13} = -1.4615384615384615$$

Solution Steps

Multiply both numerator and denominator of \frac{-3+5i}{3-2i} by the complex conjugate of the denominator, 3+2i.

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

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

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

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

$$Re(\frac{\left(-3+5i\right)\left(3+2i\right)}{13})$$

Multiply complex numbers -3+5i and 3+2i like you multiply binomials.

$$Re(\frac{-3\times 3-3\times \left(2i\right)+5i\times 3+5\times 2i^{2}}{13})$$

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

$$Re(\frac{-3\times 3-3\times \left(2i\right)+5i\times 3+5\times 2\left(-1\right)}{13})$$

Do the multiplications in -3\times 3-3\times \left(2i\right)+5i\times 3+5\times 2\left(-1\right).

$$Re(\frac{-9-6i+15i-10}{13})$$

Combine the real and imaginary parts in -9-6i+15i-10.

$$Re(\frac{-9-10+\left(-6+15\right)i}{13})$$

Do the additions in -9-10+\left(-6+15\right)i.

$$Re(\frac{-19+9i}{13})$$

Divide -19+9i by 13 to get -\frac{19}{13}+\frac{9}{13}i.

$$Re(-\frac{19}{13}+\frac{9}{13}i)$$

The real part of -\frac{19}{13}+\frac{9}{13}i is -\frac{19}{13}.

$$-\frac{19}{13}$$