Solve 3-5i/5+3i

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

Evaluate

$$-i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5-3i.

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

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

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

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

$$\frac{\left(3-5i\right)\left(5-3i\right)}{34}$$

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

$$\frac{3\times 5+3\times \left(-3i\right)-5i\times 5-5\left(-3\right)i^{2}}{34}$$

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

$$\frac{3\times 5+3\times \left(-3i\right)-5i\times 5-5\left(-3\right)\left(-1\right)}{34}$$

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

$$\frac{15-9i-25i-15}{34}$$

Combine the real and imaginary parts in 15-9i-25i-15.

$$\frac{15-15+\left(-9-25\right)i}{34}$$

Do the additions in 15-15+\left(-9-25\right)i.

$$\frac{-34i}{34}$$

Divide -34i by 34 to get -i.

$$-i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(3-5i\right)\left(5-3i\right)}{34})$$

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

$$Re(\frac{3\times 5+3\times \left(-3i\right)-5i\times 5-5\left(-3\right)i^{2}}{34})$$

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

$$Re(\frac{3\times 5+3\times \left(-3i\right)-5i\times 5-5\left(-3\right)\left(-1\right)}{34})$$

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

$$Re(\frac{15-9i-25i-15}{34})$$

Combine the real and imaginary parts in 15-9i-25i-15.

$$Re(\frac{15-15+\left(-9-25\right)i}{34})$$

Do the additions in 15-15+\left(-9-25\right)i.

$$Re(\frac{-34i}{34})$$

Divide -34i by 34 to get -i.

$$Re(-i)$$

The real part of -i is 0.

$$0$$