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**Hint:**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{align}

& =4-(-2i) \\

& =4+2i \\

\end{align}$

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{align}

& =\dfrac{8+14i}{20} \\

& \Rightarrow 12-4=8,\,\,6i+8i=14\,\,and\,\,\,-8i+8i=0 \\

\end{align}$

Now we will reduce above form into the standard form

$\begin{align}

& =\dfrac{8+14i}{20} \\

& =\dfrac{8}{20}+\dfrac{14i}{20} \\

\end{align}$

We will reduce the term a bit.

$=\dfrac{2}{5}+\dfrac{7}{10}i$

Here we get

$\begin{align}

& \dfrac{3+2i}{4-2i} \\

& =\dfrac{2}{5}+\dfrac{2i}{10} \\

\end{align}$

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.

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