Solve -4i/-1-i

$$\frac { - 4 i } { - 1 - i }$$

Evaluate

$$2+2i$$

Solution Steps

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

$$\frac{-4i\left(-1+i\right)}{\left(-1-i\right)\left(-1+i\right)}$$

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

$$\frac{-4i\left(-1+i\right)}{\left(-1\right)^{2}-i^{2}}$$

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

$$\frac{-4i\left(-1+i\right)}{2}$$

Multiply -4i times -1+i.

$$\frac{-4i\left(-1\right)-4i^{2}}{2}$$

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

$$\frac{-4i\left(-1\right)-4\left(-1\right)}{2}$$

Do the multiplications in -4i\left(-1\right)-4\left(-1\right). Reorder the terms.

$$\frac{4+4i}{2}$$

Divide 4+4i by 2 to get 2+2i.

$$2+2i$$

Real Part

$$2$$

Solution Steps

Multiply both numerator and denominator of \frac{-4i}{-1-i} by the complex conjugate of the denominator, -1+i.

$$Re(\frac{-4i\left(-1+i\right)}{\left(-1-i\right)\left(-1+i\right)})$$

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

$$Re(\frac{-4i\left(-1+i\right)}{\left(-1\right)^{2}-i^{2}})$$

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

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

Multiply -4i times -1+i.

$$Re(\frac{-4i\left(-1\right)-4i^{2}}{2})$$

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

$$Re(\frac{-4i\left(-1\right)-4\left(-1\right)}{2})$$

Do the multiplications in -4i\left(-1\right)-4\left(-1\right). Reorder the terms.

$$Re(\frac{4+4i}{2})$$

Divide 4+4i by 2 to get 2+2i.

$$Re(2+2i)$$

The real part of 2+2i is 2.

$$2$$