Solve 1+3i/3-2i

$$\frac { 1 + 3 i } { 3 - 2 i }$$

Evaluate

$$-\frac{3}{13}+\frac{11}{13}i\approx -0.230769231+0.846153846i$$

Solution Steps

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

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

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

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

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

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

Multiply complex numbers 1+3i and 3+2i like you multiply binomials.

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

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

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

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

$$\frac{3+2i+9i-6}{13}$$

Combine the real and imaginary parts in 3+2i+9i-6.

$$\frac{3-6+\left(2+9\right)i}{13}$$

Do the additions in 3-6+\left(2+9\right)i.

$$\frac{-3+11i}{13}$$

Divide -3+11i by 13 to get -\frac{3}{13}+\frac{11}{13}i.

$$-\frac{3}{13}+\frac{11}{13}i$$

Real Part

$$-\frac{3}{13} = -0.23076923076923078$$

Solution Steps

Multiply both numerator and denominator of \frac{1+3i}{3-2i} by the complex conjugate of the denominator, 3+2i.

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

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

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

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

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

Multiply complex numbers 1+3i and 3+2i like you multiply binomials.

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

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

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

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

$$Re(\frac{3+2i+9i-6}{13})$$

Combine the real and imaginary parts in 3+2i+9i-6.

$$Re(\frac{3-6+\left(2+9\right)i}{13})$$

Do the additions in 3-6+\left(2+9\right)i.

$$Re(\frac{-3+11i}{13})$$

Divide -3+11i by 13 to get -\frac{3}{13}+\frac{11}{13}i.

$$Re(-\frac{3}{13}+\frac{11}{13}i)$$

The real part of -\frac{3}{13}+\frac{11}{13}i is -\frac{3}{13}.

$$-\frac{3}{13}$$