Solveeit Logo
Login

Question

Question: The conjugate of the complex number \[\dfrac{{{{(i + 1)}^2}}}{{(1 - i)}}\] is A.\[1 - i\] B.\[1...

The conjugate of the complex number (i+1)2(1−i)\dfrac{{{{(i + 1)}^2}}}{{(1 - i)}} is
A.1−i1 - i
B.1+i1 + i
C.−1+i - 1 + i
D.−1−i - 1 - i

Explanation

Solution

Hint : We have to find the complex number of simplifications of the given expression (i+1)2(1−i)\dfrac{{{{(i + 1)}^2}}}{{(1 - i)}} . We solve this question using the concept of complex numbers and various properties of complex numbers . We firstly expand the term using the formula of sum of two terms and then simplifying the expression , we get the simplified complex number .

Complete step-by-step answer :
Given :
The given complex number is (i+1)2(1−i)\dfrac{{{{(i + 1)}^2}}}{{(1 - i)}}
Let us consider that z=(i+1)2(1−i)z = \dfrac{{{{(i + 1)}^2}}}{{(1 - i)}}
Now ,
Firstly expanding the term using the formula given as :
(a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
Using the formula , we get
z=(i2+12+2i)(1−i)z = \dfrac{{({i^2} + {1^2} + 2i)}}{{(1 - i)}}
z=(i2+1+2i)(1−i)z = \dfrac{{({i^2} + 1 + 2i)}}{{(1 - i)}}
We also know that the value of i=−1i = \sqrt { - 1}
So , we get the value of i2=−1{i^2} = - 1
Putting the value of i^2 , we get
z=(−1+1+2i)(1−i)z = \dfrac{{( - 1 + 1 + 2i)}}{{(1 - i)}}
On further simplifying , we get
z=2i(1−i)z = \dfrac{{2i}}{{(1 - i)}}
Now rationalising the term
Multiplying numerator and denominator by 1+i1 + i
z=2i×(1+i)(1−i)×(1+i)z = \dfrac{{2i \times (1 + i)}}{{(1 - i) \times \left( {1 + i} \right)}}
Using the formula (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2} and using the value of i2{i^2} , we get
z=2i×(1+i)2z = \dfrac{{2i \times (1 + i)}}{2}
Cancelling the terms , we get
z=i×(1+i)z = i \times (1 + i)
Expanding the term and putting the value of i2{i^2} , we get
z=i−1z = i - 1
We get the simplified term of the expression as z=i−1z = i - 1 .
Now , we have to find the conjugate of zz .
Conjugate of z=−1−iz = - 1 - i
Thus , The conjugate of the complex number (i+1)2(1−i)\dfrac{{{{(i + 1)}^2}}}{{(1 - i)}} is −1−i - 1 - i .
Hence , the correct option is (4)(4) .
So, the correct answer is “Option 4”.

Note : let us consider that there is a complex number zz such that z=a+ibz = a + ib . Then the conjugate of the complex number zz is given as a−iba - ib . A number of the form a+iba + ib , where aa and bb are real numbers , is called a complex number , aa is called the real part and bb is called the imaginary part of the complex number .
Every real number can be represented in terms of complex numbers but the converse is not true .
Since b2−4ac{b^2} - 4ac determines whether the quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0
If b2−4ac<0{b^2} - 4ac < 0 then the equation has imaginary roots .