Solve -4/1+i

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

Evaluate

$$-2+2i$$

Solution Steps

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

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

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

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

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

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

Multiply -4 times 1-i.

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

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

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

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

$$-2+2i$$

Real Part

$$-2$$

Solution Steps

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

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

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

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

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

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

Multiply -4 times 1-i.

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

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

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

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

$$Re(-2+2i)$$

The real part of -2+2i is -2.

$$-2$$