Solve 3-2i/2+3i

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

Evaluate

$$-i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

$$\frac{6-9i-4i-6}{13}$$

Combine the real and imaginary parts in 6-9i-4i-6.

$$\frac{6-6+\left(-9-4\right)i}{13}$$

Do the additions in 6-6+\left(-9-4\right)i.

$$\frac{-13i}{13}$$

Divide -13i by 13 to get -i.

$$-i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

$$Re(\frac{6-9i-4i-6}{13})$$

Combine the real and imaginary parts in 6-9i-4i-6.

$$Re(\frac{6-6+\left(-9-4\right)i}{13})$$

Do the additions in 6-6+\left(-9-4\right)i.

$$Re(\frac{-13i}{13})$$

Divide -13i by 13 to get -i.

$$Re(-i)$$

The real part of -i is 0.

$$0$$