Question

How do you simplify $\dfrac{1}{(1+i)}$

Answer 1

We are given a term as $\dfrac{1}{(1+i)}$ . We are asked to simplify it, we can see clearly. We can read it as $1$ is being divided by $1+i$, so we will learn about how the compared number can being divided, we will learn what we conjugate, how do we find them, we will work on some examples to get a grip, then we will solve $\dfrac{1}{(1+i)}$ by multiplying it by its conjugate and then simplifying it.

Complete step-by-step solution:
We are given a fraction as $\dfrac{1}{(1+i)}$. We can read them as $1$ is being divided by $1+i$. So, we will learn how to divide the number in complex form. To divide the number in complex form we will follow the following steps.
Step: 1 Firstly we find the conjugate of the denominator.
Step: 2 We will multiply numerator and denominator of the given fraction with the conjugate of the denominator.
Step: 3 We will then distribute the term mean. We will produce the term in numerator as well as in denominator.
Step: 4 We will simplify the power of $i$ always, remember that ${{i}^{2}}$ is given as $-1$
Step: 5 We will then combine the like terms that mean we will total infinity terms with each other and only constant with each other.
Step: 6 we will simplify our answer lastly in standard complex form. We will learn better by one example say we have $\dfrac{3+2i}{4+2i}$
We have numerator $3+2i$ and denominator as $4+2i$
So Step: 1 we will find conjugate of $4+2i$, simply the conjugate of $a+ib$ is $a-ib$ hence conjugate of $4+2i$
$\begin{aligned} & =4-(-2i) \\\ & =4+2i \\\ \end{aligned}$
Step: 2 We will multiply numerator and denominator by $4+2i$
$\Rightarrow \dfrac{3+2i}{(4-2i)}\times \dfrac{(4+2i)}{4+2i}$
Step: 3 We will multiply them in the numerator as well as in denominator
$\Rightarrow \dfrac{3+2i}{(4-2i)}\times \dfrac{(4+2i)}{4+2i}=\dfrac{12+4{{i}^{2}}+6i+8i}{16-4{{i}^{2}}+8i-8i}$
Step: 4 We simplify I we use ${{i}^{2}}=-1$
$\Rightarrow \dfrac{12-4+6i+8i}{16-(-4)+8i-8i}$
Now we combine like term and simplify
$\begin{aligned} & =\dfrac{8+14i}{20} \\\ & \Rightarrow 12-4=8,\,\,6i+8i=14\,\,and\,\,\,-8i+8i=0 \\\ \end{aligned}$
Now we will reduce above form into the standard form
$\begin{aligned} & =\dfrac{8+14i}{20} \\\ & =\dfrac{8}{20}+\dfrac{14i}{20} \\\ \end{aligned}$
We will reduce the term a bit.
$=\dfrac{2}{5}+\dfrac{7}{10}i$
Here we get
$\begin{aligned} & \dfrac{3+2i}{4-2i} \\\ & =\dfrac{2}{5}+\dfrac{2i}{10} \\\ \end{aligned}$
This is how we divide the terms in the complex number.
Now we write on our problem, we have $\dfrac{1}{(1+i)}$ conjugate of $1+i$ is $1-i$
So, we multiply numerator and denominator by $1-i$
$\dfrac{1}{(1+i)}=\dfrac{1}{1+i}\times \dfrac{1-i}{1-i}$
Simplifying by multiple we get
$\Rightarrow \dfrac{1-i}{1-i+i-{{i}^{2}}}$
Simplifying we get,
${{i}^{2}}=-1$
So, we get,
$\Rightarrow \dfrac{1-i}{1-i+i+1}$
Now we add the like term and simplify and we get
$=\dfrac{1-i}{2}$
Now we write it to standard form
$=\dfrac{1}{2}-\dfrac{i}{2}$
So, we get $\dfrac{1}{(1+i)}$ in answer as $=\dfrac{1}{2}-\dfrac{i}{2}$ into its simple form.

Note: When we have a complex term remember we cannot add the first constant with the total term. First, we cannot add variables with constant i.e. $x+2=2x$ in many similar ways $2i+2=4i$ is wrong. We can always add like terms, we do addition after simplification of another power I.e after changing ${{i}^{2}}=-1$ we need to be careful with calculation.