Solve -3i/-2-5i

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

Evaluate

$$\frac{15}{29}+\frac{6}{29}i\approx 0.517241379+0.206896552i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -2+5i.

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

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

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

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

$$\frac{-3i\left(-2+5i\right)}{29}$$

Multiply -3i times -2+5i.

$$\frac{-3i\left(-2\right)-3\times 5i^{2}}{29}$$

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

$$\frac{-3i\left(-2\right)-3\times 5\left(-1\right)}{29}$$

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

$$\frac{15+6i}{29}$$

Divide 15+6i by 29 to get \frac{15}{29}+\frac{6}{29}i.

$$\frac{15}{29}+\frac{6}{29}i$$

Real Part

$$\frac{15}{29} = 0.5172413793103449$$

Solution Steps

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

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

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

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

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

$$Re(\frac{-3i\left(-2+5i\right)}{29})$$

Multiply -3i times -2+5i.

$$Re(\frac{-3i\left(-2\right)-3\times 5i^{2}}{29})$$

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

$$Re(\frac{-3i\left(-2\right)-3\times 5\left(-1\right)}{29})$$

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

$$Re(\frac{15+6i}{29})$$

Divide 15+6i by 29 to get \frac{15}{29}+\frac{6}{29}i.

$$Re(\frac{15}{29}+\frac{6}{29}i)$$

The real part of \frac{15}{29}+\frac{6}{29}i is \frac{15}{29}.

$$\frac{15}{29}$$