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Question: Convert \( \dfrac{{i + 1}}{{\cos \dfrac{\pi }{4} - i\sin \dfrac{\pi }{4}}} \) in polar form....

Convert i+1cos⁡π4−isin⁡π4\dfrac{{i + 1}}{{\cos \dfrac{\pi }{4} - i\sin \dfrac{\pi }{4}}} in polar form.

Explanation

Solution

Hint : First we will substitute the cosine and sine functions with their values as known to us. Then we will operate accordingly to get the simplified form of the complex number. Then on achieving the simplified form, we can work accordingly to get the polar form of the given number.

Complete step-by-step answer :
Given, i+1cos⁡π4−isin⁡π4\dfrac{{i + 1}}{{\cos \dfrac{\pi }{4} - i\sin \dfrac{\pi }{4}}}
We know, cos⁡π4=sin⁡π4=12\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} .
Now, replacing this value in the above complex number, we get,
=i+112−i12= \dfrac{{i + 1}}{{\dfrac{1}{{\sqrt 2 }} - i\dfrac{1}{{\sqrt 2 }}}}
Now, taking LCM in the denominator, gives us,
=i+11−i2= \dfrac{{i + 1}}{{\dfrac{{1 - i}}{{\sqrt 2 }}}}
Simplifying the expression, this can be written as,
=2(i+1)1−i= \dfrac{{\sqrt 2 \left( {i + 1} \right)}}{{1 - i}}
Multiplying the numerator and denominator by (1+i)\left( {1 + i} \right) , we get,
=2(1+i)(1+i)(1−i)(1+i)= \dfrac{{\sqrt 2 \left( {1 + i} \right)\left( {1 + i} \right)}}{{\left( {1 - i} \right)\left( {1 + i} \right)}}
Using the algebraic identity (a−b)(a+b)=a2−b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} ,
=2(1+i)2(1−i2)= \dfrac{{\sqrt 2 {{\left( {1 + i} \right)}^2}}}{{\left( {1 - {i^2}} \right)}}
We know, i=−1i = \sqrt { - 1}
⇒i2=−1\Rightarrow {i^2} = - 1
Therefore, substituting this value, we get,
=2(1+i)2[1−(−1)]= \dfrac{{\sqrt 2 {{\left( {1 + i} \right)}^2}}}{{\left[ {1 - \left( { - 1} \right)} \right]}}
In the numerator, using (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} , we get,
=2(1+2i+i2)[1+1]= \dfrac{{\sqrt 2 \left( {1 + 2i + {i^2}} \right)}}{{\left[ {1 + 1} \right]}}
Again substituting the values, we get,
=2(1+2i+(−1))2= \dfrac{{\sqrt 2 \left( {1 + 2i + \left( { - 1} \right)} \right)}}{2}
=2(1+2i−1)2= \dfrac{{\sqrt 2 \left( {1 + 2i - 1} \right)}}{2}
Now, simplifying, we get,
=2×(2i)2= \dfrac{{\sqrt 2 \times (2i)}}{2}
=2i= \sqrt 2 i
Now, we can clearly see that the modulus of the complex number is 02+(2)2=2\sqrt {{0^2} + {{\left( {\sqrt 2 } \right)}^2}} = \sqrt 2 . So, we can write the complex number as,
=2(0+i)= \sqrt 2 \left( {0 + i} \right)
Now, we know the values of sine and cosine of π2\dfrac{\pi }{2} angle as 00 and 11 respectively. So, we get,
=2(cos⁡π2+isin⁡π2)= \sqrt 2 \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)
Therefore, the polar form of i+1cos⁡π4−isin⁡π4\dfrac{{i + 1}}{{\cos \dfrac{\pi }{4} - i\sin \dfrac{\pi }{4}}} is 2(cos⁡π2+isin⁡π2)\sqrt 2 \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right) .

Note : Conventionally we find the polar form of a complex number such as a+iba + ib , by finding the modulus value of the number as r=a2+b2r = \sqrt {{a^2} + {b^2}} and finding the angle ϕ=tan⁡−1∣ba∣\phi = {\tan ^{ - 1}}\left| {\dfrac{b}{a}} \right| and then finding in which quadrant the number lies depending on the sign of the numbers aa and bb , and then finding the angle θ\theta of the polar number which is quadrant specified whereas ϕ\phi is not quadrant specified. Then we find the polar form of the complex number as r(cos⁡θ+isin⁡θ)r\left( {\cos \theta + i\sin \theta } \right) .