Solveeit Logo
Login

Question

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

If c2+s2=1{{c}^{2}}+{{s}^{2}}=1 , then find the value of 1+c+is1+c−is\dfrac{1+c+is}{1+c-is}:
(a) c + is
(b) c – is
(c) s + ic
(d) s – ic

Explanation

Solution

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 c2+s2=1{{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 (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab , we are going to use this formula for calculating the value of z2{{z}^{2}}, where z can any complex number.
Another formula that we are going to use is (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} and i2=−1{{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.
⇒(1+c+is1+c−is)(1+c+is1+c+is) ⇒(1+c+is)2(1+c)2−(is)2 \begin{aligned} & \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{aligned}
Now we will use (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab to expand,
We know that i2=−1{{i}^{2}}=-1 , using this we get,
⇒(1+c)2+(is)2+2s(1+c)i(1+c)2+s2\Rightarrow \dfrac{{{(1+c)}^{2}}+{{\left( is \right)}^{2}}+2s(1+c)i}{{{(1+c)}^{2}}+{{s}^{2}}}
We know that i2=−1{{i}^{2}}=-1 , using this we get,
⇒(1+c)2−s2+2s(1+c)i1+c2+2c+s2\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 c2+s2=1{{c}^{2}}+{{s}^{2}}=1 to solve this equation.
Now using c2+s2=1{{c}^{2}}+{{s}^{2}}=1 in (1+c)2−s2+2s(1+c)i1+c2+2c+s2\dfrac{{{(1+c)}^{2}}-{{s}^{2}}+2s(1+c)i}{1+{{c}^{2}}+2c+{{s}^{2}}} we get,
⇒(1+c)2−(1−c2)+2s(1+c)i2+2c ⇒(1+c)2+c2−1+2s(1+c)i2+2c ⇒(1+c)2+(c−1)(c+1)+2s(1+c)i2(c+1) \begin{aligned} & \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{aligned}
Now taking (c + 1) common and canceling from both numerator and denominator we get,
⇒(1+c)+(c−1)+2si2 ⇒2c+2si2 ⇒c+is \begin{aligned} & \Rightarrow \dfrac{(1+c)+(c-1)+2si}{2} \\\ & \Rightarrow \dfrac{2c+2si}{2} \\\ & \Rightarrow c+is \\\ \end{aligned}
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 c2+s2=1{{c}^{2}}+{{s}^{2}}=1 to solve our question, so analyse the answer carefully to understand when and where to use this relation.