Question

If $z$ is a point on the Argand plane such that \[\left| z-1 \right|=1\] , then $\dfrac{z-2}{z}$ is equal to
A. $\tan (\arg (z-1)$
B. $\cot (\arg (z-1))$
C. $i\tan (\arg (z-1))$
D. None of these

Answer 1

Hint: We need to find the value of $\dfrac{z-2}{z}$ given that \[\left| z-1 \right|=1\] , where $z$ is a point on the Argand plane. Since, \[\left| z-1 \right|=1\] we will take $z-1=\cos \theta +i\sin \theta $ . By adding $-1$ on both the sides, we will get $z-2$ . By applying suitable trigonometric properties we will get an equation for $z-2$ . Now, from $z-1=\cos \theta +i\sin \theta $ , we will get the equation for $z$ . By using suitable trigonometric properties, we will get an equation similar to the one for $z-2$ . Now divide these equations to get value of $\dfrac{z-2}{z}$ .

Complete step by step answer:
In order to find the value of $\dfrac{z-2}{z}$ given that \[\left| z-1 \right|=1\] , where $z$ is a point on the Argand plane, let us assume that
  $z-1=\cos \theta +i\sin \theta ...(a)$ .
Thus \[\left| z-1 \right|=\sqrt{{{\cos }^{2}}\theta +\sin {{\theta }^{2}}}=1\]
Hence the given condition \[\left| z-1 \right|=1\].
So $z-2$ can be written by adding $-1$ on both the sides of equation $(a)$ .
$\Rightarrow z-1-1=\cos \theta +i\sin \theta -1$
By adding, we get
$z-2=\cos \theta +i\sin \theta -1...(i)$
We know that $\cos 2\theta =1-2{{\sin }^{2}}\theta $ .
Hence, $\cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2}$
Substituting this in $(i)$ , we get
$z-2=-2{{\sin }^{2}}\dfrac{\theta }{2}+i\sin \theta ...(ii)$
We know that $\sin 2\theta =2\sin \theta \cos \theta $
Hence, $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
Thus, equation $(ii)$ can be written as
$z-2=-2{{\sin }^{2}}\dfrac{\theta }{2}+2i\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
By taking $2\sin \dfrac{\theta }{2}$ as common, we will get
\[z-2=2\sin \dfrac{\theta }{2}\left( -\sin \dfrac{\theta }{2}+i\cos \dfrac{\theta }{2} \right)\]
Since ${{i}^{2}}=-1$ , the above equation can be written as
\[z-2=2\sin \dfrac{\theta }{2}\left( i\times i\sin \dfrac{\theta }{2}+i\cos \dfrac{\theta }{2} \right)\]
Taking $i$ outside from RHS gives the following equation, we will get
\[z-2=2i\sin \dfrac{\theta }{2}\left( i\sin \dfrac{\theta }{2}+\cos \dfrac{\theta }{2} \right)...(b)\]
Now, equation $(a)$ can be written as
$z=\cos \theta +i\sin \theta +1...(iii)$
We know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$ .
Hence, $\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}-1$
Thus equation $(iii)$ can be written as
$z=2{{\cos }^{2}}\dfrac{\theta }{2}-1+i\sin \theta +1$
By solving, we get
$z=2{{\cos }^{2}}\dfrac{\theta }{2}+i\sin \theta $
We know that $\sin 2\theta =2\sin \theta \cos \theta $
Hence, $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
Thus, $z=2{{\cos }^{2}}\dfrac{\theta }{2}+2i\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
By taking $2\cos \dfrac{\theta }{2}$ as common, we will get
$z=2\cos \dfrac{\theta }{2}\left( \cos \dfrac{\theta }{2}+i\sin \dfrac{\theta }{2} \right)...(c)$
Dividing equation $(b)\text{ by }(c)$ , we get
\[\dfrac{z-2}{z}=\dfrac{2i\sin \dfrac{\theta }{2}\left( i\sin \dfrac{\theta }{2}+\cos \dfrac{\theta }{2} \right)}{2\cos \dfrac{\theta }{2}\left( \cos \dfrac{\theta }{2}+i\sin \dfrac{\theta }{2} \right)}\]
Cancelling the common term gives
$\dfrac{z-2}{z}=\dfrac{i\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}}$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ .
Thus, $\dfrac{z-2}{z}=i\tan \dfrac{\theta }{2}$
Now, $\arg (z-1)=\dfrac{\theta }{2}$
Thus, \[\dfrac{z-2}{z}=i\tan (\arg (z-1))\]
Hence, the correct option is C.

Note:
We used $z-1=\cos \theta +i\sin \theta $ since \[\left| z-1 \right|=1\] . If any such condition is not given, we could use only $z=\cos \theta +i\sin \theta $ . Be careful with the value of ${{i}^{2}}=-1$ . Students can make error in this by considering ${{i}^{2}}=1$ . Trigonometric properties should be thorough as this problem is mainly solved using that.