Solve -2+i/-1-8i

$$\frac { - 2 + i } { - 1 - 8 i }$$

Evaluate

$$-\frac{6}{65}-\frac{17}{65}i\approx -0.092307692-0.261538462i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -1+8i.

$$\frac{\left(-2+i\right)\left(-1+8i\right)}{\left(-1-8i\right)\left(-1+8i\right)}$$

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

$$\frac{\left(-2+i\right)\left(-1+8i\right)}{\left(-1\right)^{2}-8^{2}i^{2}}$$

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

$$\frac{\left(-2+i\right)\left(-1+8i\right)}{65}$$

Multiply complex numbers -2+i and -1+8i like you multiply binomials.

$$\frac{-2\left(-1\right)-2\times \left(8i\right)-i+8i^{2}}{65}$$

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

$$\frac{-2\left(-1\right)-2\times \left(8i\right)-i+8\left(-1\right)}{65}$$

Do the multiplications in -2\left(-1\right)-2\times \left(8i\right)-i+8\left(-1\right).

$$\frac{2-16i-i-8}{65}$$

Combine the real and imaginary parts in 2-16i-i-8.

$$\frac{2-8+\left(-16-1\right)i}{65}$$

Do the additions in 2-8+\left(-16-1\right)i.

$$\frac{-6-17i}{65}$$

Divide -6-17i by 65 to get -\frac{6}{65}-\frac{17}{65}i.

$$-\frac{6}{65}-\frac{17}{65}i$$

Real Part

$$-\frac{6}{65} = -0.09230769230769231$$

Solution Steps

Multiply both numerator and denominator of \frac{-2+i}{-1-8i} by the complex conjugate of the denominator, -1+8i.

$$Re(\frac{\left(-2+i\right)\left(-1+8i\right)}{\left(-1-8i\right)\left(-1+8i\right)})$$

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

$$Re(\frac{\left(-2+i\right)\left(-1+8i\right)}{\left(-1\right)^{2}-8^{2}i^{2}})$$

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

$$Re(\frac{\left(-2+i\right)\left(-1+8i\right)}{65})$$

Multiply complex numbers -2+i and -1+8i like you multiply binomials.

$$Re(\frac{-2\left(-1\right)-2\times \left(8i\right)-i+8i^{2}}{65})$$

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

$$Re(\frac{-2\left(-1\right)-2\times \left(8i\right)-i+8\left(-1\right)}{65})$$

Do the multiplications in -2\left(-1\right)-2\times \left(8i\right)-i+8\left(-1\right).

$$Re(\frac{2-16i-i-8}{65})$$

Combine the real and imaginary parts in 2-16i-i-8.

$$Re(\frac{2-8+\left(-16-1\right)i}{65})$$

Do the additions in 2-8+\left(-16-1\right)i.

$$Re(\frac{-6-17i}{65})$$

Divide -6-17i by 65 to get -\frac{6}{65}-\frac{17}{65}i.

$$Re(-\frac{6}{65}-\frac{17}{65}i)$$

The real part of -\frac{6}{65}-\frac{17}{65}i is -\frac{6}{65}.

$$-\frac{6}{65}$$