Solve 2+3i/4-5i

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

Evaluate

$$-\frac{7}{41}+\frac{22}{41}i\approx -0.170731707+0.536585366i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\frac{-7+22i}{41}$$

Divide -7+22i by 41 to get -\frac{7}{41}+\frac{22}{41}i.

$$-\frac{7}{41}+\frac{22}{41}i$$

Real Part

$$-\frac{7}{41} = -0.17073170731707318$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$Re(\frac{-7+22i}{41})$$

Divide -7+22i by 41 to get -\frac{7}{41}+\frac{22}{41}i.

$$Re(-\frac{7}{41}+\frac{22}{41}i)$$

The real part of -\frac{7}{41}+\frac{22}{41}i is -\frac{7}{41}.

$$-\frac{7}{41}$$