Question

If ${{c}^{2}}+{{s}^{2}}=1$ , then find the value of $\dfrac{1+c+is}{1+c-is}$:
(a) c + is
(b) c – is
(c) s + ic
(d) s – ic

Answer 1

Hint: First we will solve the denominator separately and write in the form of x + iy, then we will rationalize the denominator by multiplying x – iy in both numerator and denominator and then we have to rearrange some terms to make it in the form of a + ib. Then we will use ${{c}^{2}}+{{s}^{2}}=1$ to find out the value.

Complete step-by-step answer:
Let’s first rationalize the denominator,
Rationalizing the denominator means multiplying it with some numbers does not make it into an integer.
The formula for ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we are going to use this formula for calculating the value of ${{z}^{2}}$, where z can any complex number.
Another formula that we are going to use is $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ and ${{i}^{2}}=-1$ ,
Let’s use all these formulas to solve this question.
The conjugate of ( a + ib ) is ( a – ib ).
We are multiplying by (1 + c + is) because whenever we multiply a complex number by it’s conjugate we get a real number and that is our objective here.
$\begin{align}
  & \Rightarrow \left( \dfrac{1+c+is}{1+c-is} \right)\left( \dfrac{1+c+is}{1+c+is} \right) \\
 & \Rightarrow \dfrac{{{(1+c+is)}^{2}}}{{{(1+c)}^{2}}-{{\left( is \right)}^{2}}} \\
\end{align}$
Now we will use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to expand,
We know that ${{i}^{2}}=-1$ , using this we get,
$\Rightarrow \dfrac{{{(1+c)}^{2}}+{{\left( is \right)}^{2}}+2s(1+c)i}{{{(1+c)}^{2}}+{{s}^{2}}}$
We know that ${{i}^{2}}=-1$ , using this we get,
$\Rightarrow \dfrac{{{(1+c)}^{2}}-{{s}^{2}}+2s(1+c)i}{1+{{c}^{2}}+2c+{{s}^{2}}}$
Now we have rationalized the denominator and after that we will use this relation ${{c}^{2}}+{{s}^{2}}=1$ to solve this equation.
Now using ${{c}^{2}}+{{s}^{2}}=1$ in $\dfrac{{{(1+c)}^{2}}-{{s}^{2}}+2s(1+c)i}{1+{{c}^{2}}+2c+{{s}^{2}}}$ we get,
$\begin{align}
  & \Rightarrow \dfrac{{{(1+c)}^{2}}-\left( 1-{{c}^{2}} \right)+2s(1+c)i}{2+2c} \\
 & \Rightarrow \dfrac{{{(1+c)}^{2}}+{{c}^{2}}-1+2s(1+c)i}{2+2c} \\
 & \Rightarrow \dfrac{{{(1+c)}^{2}}+\left( c-1 \right)\left( c+1 \right)+2s(1+c)i}{2\left( c+1 \right)} \\
\end{align}$
Now taking (c + 1) common and canceling from both numerator and denominator we get,
$\begin{align}
  & \Rightarrow \dfrac{(1+c)+(c-1)+2si}{2} \\
 & \Rightarrow \dfrac{2c+2si}{2} \\
 & \Rightarrow c+is \\
\end{align}$
So, after solving and using all the information that was given we get the value of the equation as c + is.
Hence, the option (a) is correct.

Note: There are some important concepts which are required in this question and one is the conjugate of a complex number, the conjugate of ( a + ib ) is ( a – ib ). Another very important step where most of the students get stuck is how to use the given relation ${{c}^{2}}+{{s}^{2}}=1$ to solve our question, so analyse the answer carefully to understand when and where to use this relation.