How to solve - 2/(1 + i)

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve - 2/(1 + i) | Equatix

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Question

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

Evaluate

$-1+i$

Solution Steps

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

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

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

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

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

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

Multiply $-2$ times $1-i$.

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

Do the multiplications in $-2-2\left(-i\right)$.

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

Divide $-2+2i$ by $2$ to get $-1+i$.

$$-1+i$$

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Real Part

$-1$

Solution Steps

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

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

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

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

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

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

Multiply $-2$ times $1-i$.

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

Do the multiplications in $-2-2\left(-i\right)$.

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

Divide $-2+2i$ by $2$ to get $-1+i$.

$$Re(-1+i)$$

The real part of $-1+i$ is $-1$.

$$-1$$

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Basic Math

Pre-Algebra

Algebra

Trigonometry

Calculus

Precalculus

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Example

Question

Evaluate

Solution Steps

Real Part

Solution Steps