Solve -2i/-6-5i

$$\frac { - 2 i } { - 6 - 5 i }$$

Evaluate

$$\frac{10}{61}+\frac{12}{61}i\approx 0.163934426+0.196721311i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -6+5i.

$$\frac{-2i\left(-6+5i\right)}{\left(-6-5i\right)\left(-6+5i\right)}$$

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

$$\frac{-2i\left(-6+5i\right)}{\left(-6\right)^{2}-5^{2}i^{2}}$$

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

$$\frac{-2i\left(-6+5i\right)}{61}$$

Multiply -2i times -6+5i.

$$\frac{-2i\left(-6\right)-2\times 5i^{2}}{61}$$

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

$$\frac{-2i\left(-6\right)-2\times 5\left(-1\right)}{61}$$

Do the multiplications in -2i\left(-6\right)-2\times 5\left(-1\right). Reorder the terms.

$$\frac{10+12i}{61}$$

Divide 10+12i by 61 to get \frac{10}{61}+\frac{12}{61}i.

$$\frac{10}{61}+\frac{12}{61}i$$

Real Part

$$\frac{10}{61} = 0.16393442622950818$$

Solution Steps

Multiply both numerator and denominator of \frac{-2i}{-6-5i} by the complex conjugate of the denominator, -6+5i.

$$Re(\frac{-2i\left(-6+5i\right)}{\left(-6-5i\right)\left(-6+5i\right)})$$

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

$$Re(\frac{-2i\left(-6+5i\right)}{\left(-6\right)^{2}-5^{2}i^{2}})$$

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

$$Re(\frac{-2i\left(-6+5i\right)}{61})$$

Multiply -2i times -6+5i.

$$Re(\frac{-2i\left(-6\right)-2\times 5i^{2}}{61})$$

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

$$Re(\frac{-2i\left(-6\right)-2\times 5\left(-1\right)}{61})$$

Do the multiplications in -2i\left(-6\right)-2\times 5\left(-1\right). Reorder the terms.

$$Re(\frac{10+12i}{61})$$

Divide 10+12i by 61 to get \frac{10}{61}+\frac{12}{61}i.

$$Re(\frac{10}{61}+\frac{12}{61}i)$$

The real part of \frac{10}{61}+\frac{12}{61}i is \frac{10}{61}.

$$\frac{10}{61}$$