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Question: If \[\dfrac{z-1}{z+1}\] is purely imaginary then what would be the locus of \(z\) A) \[x^{2}+y^{2}...

If z−1z+1\dfrac{z-1}{z+1} is purely imaginary then what would be the locus of zz
A) x2+y2=1x^{2}+y^{2}=1
B) x2+y2=4x^{2}+y^{2}=4
C) x+y=11x+y=11
D) 2xy=x−y2xy=x-y

Explanation

Solution

Hint: In this question it is given that If z−1z+1\dfrac{z-1}{z+1} is purely imaginary then we have to find the locus of zz, where zz is a complex number. So to find the solution we have to consider z=x+iy, where x and y be any two real number and after putting the value of ‘z’ in the given expression z−1z+1\dfrac{z-1}{z+1} we will able to find the solution.
Complete step-by-step solution:
Let,
S=z−1z+1S=\dfrac{z-1}{z+1}
=x+iy−1x+iy+1=\dfrac{x+iy-1}{x+iy+1}
=(x−1)+iy(x+1)+iy=\dfrac{\left( x-1\right) +iy}{\left( x+1\right) +iy}
Now multiplying numerator and denominator by (x+1)−iy(x+1)-iy, we get,
S=(x−1)+iy(x+1)+iy⋅(x+1)−iy(x+1)−iyS=\dfrac{\left( x-1\right) +iy}{\left( x+1\right) +iy} \cdot \dfrac{\left( x+1\right) -iy}{\left( x+1\right) -iy}
Now since, as we know that (a+b)(a−b)=a2−b2\left( a+b\right) \left( a-b\right) =a^{2}-b^{2},
So by using the above identity where a=x+1a=x+1 and b=iyb=iy, we get,
S=\dfrac{\left\\{ \left( x-1\right) +iy\right\\} \left\\{ \left( x+1\right) -iy\right\\} }{\left( x+1\right)^{2} -\left( iy\right)^{2} }
=(x−1)(x+1)+iy(x+1)−iy(x−1)−(iy)2(x+1)2−i2y2=\dfrac{\left( x-1\right) \left( x+1\right) +iy\left( x+1\right) -iy\left( x-1\right) -\left( iy\right)^{2} }{\left( x+1\right)^{2} -i^{2}y^{2}}
=x2−1+iyx+iy−iyx+iy−i2y2(x+1)2−i2y2=\dfrac{x^{2}-1+iy x+iy-iy x+iy-i^{2}y^{2}}{\left( x+1\right)^{2} -i^{2}y^{2}}
Since, ‘ii’ is the square root of -1, therefore i2=−1i^{2}=-1,
S=x2−1+2iy−(−1)y2(x+1)2−(−1)y2S=\dfrac{x^{2}-1+2iy-\left( -1\right) y^{2}}{\left( x+1\right)^{2} -\left( -1\right) y^{2}}
=x2−1+2iy+y2(x+1)2+y2=\dfrac{x^{2}-1+2iy+y^{2}}{\left( x+1\right)^{2} +y^{2}}
=(x2−1+y2)+i(2y)(x+1)2+y2=\dfrac{(x^{2}-1+y^{2})+i(2y)}{\left( x+1\right)^{2} +y^{2}}
=(x2−1+y2)(x+1)2+y2+i⋅2y(x+1)2+y2=\dfrac{(x^{2}-1+y^{2})}{\left( x+1\right)^{2} +y^{2}} +i\cdot \dfrac{2y}{\left( x+1\right)^{2} +y^{2}}
Now since, in the question it is given that S=z−1z+1S=\dfrac{z-1}{z+1} is purely imaginary, therefore its real part will be zero,
i.e, Re(z−1z+1)=0\left( \dfrac{z-1}{z+1} \right) =0
Which implies,
x2−1+y2(x+1)2+y2=0\dfrac{x^{2}-1+y^{2}}{\left( x+1\right)^{2} +y^{2}} =0
⇒x2−1+y2=0\Rightarrow x^{2}-1+y^{2}=0
⇒x2+y2=1\Rightarrow x^{2}+y^{2}=1
Therefore, the locus of ‘z’ is x2+y2=1x^{2}+y^{2}=1.
Hence, the correct option is option A.
Note: While solving this kind of question you need to know that, in mathematics ‘z’ defines the complex number which can be written as z=x+iy, where ‘x’ is the real part and ‘y’ is the imaginary part and ‘i’ is the unit imaginary number whose value is i=−1i=\sqrt{-1}.