Solve 2+3i/5-4i

$$\frac{ 2+3i }{ 5-4i }$$

Evaluate

$$-\frac{2}{41}+\frac{23}{41}i\approx -0.048780488+0.56097561i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5+4i.

$$\frac{\left(2+3i\right)\left(5+4i\right)}{\left(5-4i\right)\left(5+4i\right)}$$

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

$$\frac{\left(2+3i\right)\left(5+4i\right)}{5^{2}-4^{2}i^{2}}$$

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

$$\frac{\left(2+3i\right)\left(5+4i\right)}{41}$$

Multiply complex numbers 2+3i and 5+4i like you multiply binomials.

$$\frac{2\times 5+2\times \left(4i\right)+3i\times 5+3\times 4i^{2}}{41}$$

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

$$\frac{2\times 5+2\times \left(4i\right)+3i\times 5+3\times 4\left(-1\right)}{41}$$

Do the multiplications in 2\times 5+2\times \left(4i\right)+3i\times 5+3\times 4\left(-1\right).

$$\frac{10+8i+15i-12}{41}$$

Combine the real and imaginary parts in 10+8i+15i-12.

$$\frac{10-12+\left(8+15\right)i}{41}$$

Do the additions in 10-12+\left(8+15\right)i.

$$\frac{-2+23i}{41}$$

Divide -2+23i by 41 to get -\frac{2}{41}+\frac{23}{41}i.

$$-\frac{2}{41}+\frac{23}{41}i$$

Real Part

$$-\frac{2}{41} = -0.04878048780487805$$

Solution Steps

Multiply both numerator and denominator of \frac{2+3i}{5-4i} by the complex conjugate of the denominator, 5+4i.

$$Re(\frac{\left(2+3i\right)\left(5+4i\right)}{\left(5-4i\right)\left(5+4i\right)})$$

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

$$Re(\frac{\left(2+3i\right)\left(5+4i\right)}{5^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{\left(2+3i\right)\left(5+4i\right)}{41})$$

Multiply complex numbers 2+3i and 5+4i like you multiply binomials.

$$Re(\frac{2\times 5+2\times \left(4i\right)+3i\times 5+3\times 4i^{2}}{41})$$

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

$$Re(\frac{2\times 5+2\times \left(4i\right)+3i\times 5+3\times 4\left(-1\right)}{41})$$

Do the multiplications in 2\times 5+2\times \left(4i\right)+3i\times 5+3\times 4\left(-1\right).

$$Re(\frac{10+8i+15i-12}{41})$$

Combine the real and imaginary parts in 10+8i+15i-12.

$$Re(\frac{10-12+\left(8+15\right)i}{41})$$

Do the additions in 10-12+\left(8+15\right)i.

$$Re(\frac{-2+23i}{41})$$

Divide -2+23i by 41 to get -\frac{2}{41}+\frac{23}{41}i.

$$Re(-\frac{2}{41}+\frac{23}{41}i)$$

The real part of -\frac{2}{41}+\frac{23}{41}i is -\frac{2}{41}.

$$-\frac{2}{41}$$