Question

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

Answer 1

Hint: In this question it is given that If $$\dfrac{z-1}{z+1}$$ is purely imaginary then we have to find the locus of $z$, where $z$ 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 $$\dfrac{z-1}{z+1}$$ we will able to find the solution.
Complete step-by-step solution:
Let,
$$S=\dfrac{z-1}{z+1}$$
$$=\dfrac{x+iy-1}{x+iy+1}$$
$$=\dfrac{\left( x-1\right) +iy}{\left( x+1\right) +iy}$$
Now multiplying numerator and denominator by $(x+1)-iy$, we get,
$$S=\dfrac{\left( x-1\right) +iy}{\left( x+1\right) +iy} \cdot \dfrac{\left( x+1\right) -iy}{\left( x+1\right) -iy}$$
Now since, as we know that $$\left( a+b\right) \left( a-b\right) =a^{2}-b^{2}$$,
So by using the above identity where $a=x+1$ and $b=iy$, we get,
$$S=\dfrac{\left\{ \left( x-1\right) +iy\right\} \left\{ \left( x+1\right) -iy\right\} }{\left( x+1\right)^{2} -\left( iy\right)^{2} }$$
$$=\dfrac{\left( x-1\right) \left( x+1\right) +iy\left( x+1\right) -iy\left( x-1\right) -\left( iy\right)^{2} }{\left( x+1\right)^{2} -i^{2}y^{2}}$$
$$=\dfrac{x^{2}-1+iy x+iy-iy x+iy-i^{2}y^{2}}{\left( x+1\right)^{2} -i^{2}y^{2}}$$
Since, ‘$i$’ is the square root of -1, therefore $i^{2}=-1$,
$$S=\dfrac{x^{2}-1+2iy-\left( -1\right) y^{2}}{\left( x+1\right)^{2} -\left( -1\right) y^{2}}$$
$$=\dfrac{x^{2}-1+2iy+y^{2}}{\left( x+1\right)^{2} +y^{2}}$$
$$=\dfrac{(x^{2}-1+y^{2})+i(2y)}{\left( x+1\right)^{2} +y^{2}}$$
$$=\dfrac{(x^{2}-1+y^{2})}{\left( x+1\right)^{2} +y^{2}} +i\cdot \dfrac{2y}{\left( x+1\right)^{2} +y^{2}}$$
Now since, in the question it is given that $$S=\dfrac{z-1}{z+1}$$ is purely imaginary, therefore its real part will be zero,
i.e, Re$$\left( \dfrac{z-1}{z+1} \right) =0$$
Which implies,
$$\dfrac{x^{2}-1+y^{2}}{\left( x+1\right)^{2} +y^{2}} =0$$
$$\Rightarrow x^{2}-1+y^{2}=0$$
$$\Rightarrow x^{2}+y^{2}=1$$
Therefore, the locus of ‘z’ is $x^{2}+y^{2}=1$.
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 $i=\sqrt{-1}$.