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Question: The conjugate of a complex number is \(\dfrac{1}{i-1}\) , then that complex number is (a)\(\dfrac{...

The conjugate of a complex number is 1i−1\dfrac{1}{i-1} , then that complex number is
(a)−1i−1\dfrac{-1}{i-1}
(b)1i+1\dfrac{1}{i+1}
(c)−1i+1\dfrac{-1}{i+1}
(d)1i−1\dfrac{1}{i-1}

Explanation

Solution

Hint: First rationalize the given term, then apply the concept of conjugate of this rationalized complex number then you will get the conjugate. Try to manipulate and convert it into anyone of the options formats. That will be your required result.

Complete step-by-step answer:
Given complex number in the question can be written as:
1i−1\dfrac{1}{i-1}
Let us assume this complex number to be denoted by K. By writing equation of K with its value, we get it as:
K=1i−1K=\dfrac{1}{i-1}
Now we need to rationalize K. The process followed is given by multiplying and dividing with (i +1) in them “K”, we get it as
K=1i−1×(i+1)(i+1)K=\dfrac{1}{i-1}\times \dfrac{\left( i+1 \right)}{\left( i+1 \right)}
By using distributive law a(b + c) = ab + ac, we get K value as:
K=i+1i2−i+i−1K=\dfrac{i+1}{{{i}^{2}}-i+i-1}
By cancelling common terms, we get the value of K as
K=i+1i2−1K=\dfrac{i+1}{{{i}^{2}}-1}
By substituting value of square of i as -1, we get it as
K=i+1−1−1K=\dfrac{i+1}{-1-1}
By simplifying the above term, we can write it in form of:
K=−i2−12...............(1)K=\dfrac{-i}{2}-\dfrac{1}{2}...............\left( 1 \right)
Conjugate of a complex number: The complex number whose real parts are equal, imaginary parts are equal in magnitude but opposite in signs are said to be complex conjugate of each other. We get them by simply reversing the sign of the imaginary part of the original complex number. So, by reversing imaginary part of K, we get conjugate of K as
Conjugate of K=−12+i2K=\dfrac{-1}{2}+\dfrac{i}{2}
By multiplying and dividing with i + 1, we can write it as
Conjugate of K=(i−1)(i+1)2(i+1)K=\dfrac{\left( i-1 \right)\left( i+1 \right)}{2\left( i+1 \right)}
By using distributive law a(b + c) = ab + ac, we can write it as
Conjugate of K=i2−i+i−12(i+1)=i2−12(i+1)=−22(i+1)K=\dfrac{{{i}^{2}}-i+i-1}{2\left( i+1 \right)}=\dfrac{{{i}^{2}}-1}{2\left( i+1 \right)}=\dfrac{-2}{2\left( i+1 \right)}
By simplifying, we can write it as following equation:
Conjugate of K=−1i+1K=\dfrac{-1}{i+1}
Therefore option (c) is the correct answer.
Note: You can directly reverse the sign in the first step to get the result in a single step. Anyways you will get the same result. Be careful with signs while doing rationalisation, as if you write 1−i21-{{i}^{2}} instead of i2−1{{i}^{2}}-1 , the whole sign of the result might change. So, do that step carefully.