Solve 3+2i/5-3i

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

Evaluate

$$\frac{9}{34}+\frac{19}{34}i\approx 0.264705882+0.558823529i$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$\frac{3\times 5+3\times \left(3i\right)+2i\times 5+2\times 3\left(-1\right)}{34}$$

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

$$\frac{15+9i+10i-6}{34}$$

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

$$\frac{15-6+\left(9+10\right)i}{34}$$

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

$$\frac{9+19i}{34}$$

Divide 9+19i by 34 to get \frac{9}{34}+\frac{19}{34}i.

$$\frac{9}{34}+\frac{19}{34}i$$

Real Part

$$\frac{9}{34} = 0.2647058823529412$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(\frac{3\times 5+3\times \left(3i\right)+2i\times 5+2\times 3\left(-1\right)}{34})$$

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

$$Re(\frac{15+9i+10i-6}{34})$$

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

$$Re(\frac{15-6+\left(9+10\right)i}{34})$$

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

$$Re(\frac{9+19i}{34})$$

Divide 9+19i by 34 to get \frac{9}{34}+\frac{19}{34}i.

$$Re(\frac{9}{34}+\frac{19}{34}i)$$

The real part of \frac{9}{34}+\frac{19}{34}i is \frac{9}{34}.

$$\frac{9}{34}$$