Question

If amp $\dfrac{z-1}{z+1}$ = $\dfrac{\pi }{3}$ , then z represents
(a) a straight line
(b) a circle
(c) a pair of lines
(d) none of these

Answer 1

Hint : To solve this question, we will first assume a value for complex number z. Then solve the given expression so that we can separate the real and imaginary part. We will have to rationalise the expression. We know that amp(z) = ${{\tan }^{-1}}\left( \dfrac{\text{imaginary}\ \text{part}}{\text{real}\ \text{part}} \right)$ . Thus, using this we will find the resultant equation and based on the equation, we can say which type of graph is traced by z.

Complete step-by-step answer :
We know that z is a complex number with a real part and an imaginary part.
Let us assume that z = x + iy, where i = $\sqrt{-1}$
Now, we will find the value of $\dfrac{z-1}{z+1}$ .
We will substitute z = x + iy in $\dfrac{z-1}{z+1}$ .
$\Rightarrow \dfrac{z-1}{z+1}=\dfrac{x-1+iy}{x+1+iy}$
Now, we need to remove i from the denominator. To do this, we will multiply the denominator by its conjugate.
Thus, the conjugate of x + 1 + iy is x + 1 – iy.

& \Rightarrow \dfrac{z-1}{z+1}=\dfrac{x-1+iy}{x+1+iy}\times \dfrac{x+1-iy}{x+1-iy} \\\ & \Rightarrow \dfrac{z-1}{z+1}=\dfrac{{{x}^{2}}+x-ixy-x-1+iy+ixy+iy-{{i}^{2}}{{y}^{2}}}{{{\left( x+1 \right)}^{2}}-{{i}^{2}}{{y}^{2}}} \\\ \end{aligned}$$ We know that $ {{i}^{2}} $ = ─1. $$\Rightarrow \dfrac{z-1}{z+1}=\dfrac{{{x}^{2}}+x-ixy-x-1+iy+ixy+iy+{{y}^{2}}}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}$$ Now, we will separate the real part and the imaginary part of the complex number. $$\Rightarrow \dfrac{z-1}{z+1}=\dfrac{{{x}^{2}}-1+{{y}^{2}}}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}+i\dfrac{2y}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}$$ Now, as we know the amp(z) = $ {{\tan }^{-1}}\left( \dfrac{\text{imaginary}\ \text{part}}{\text{real}\ \text{part}} \right) $ Thus, amp = $ {{\tan }^{-1}}\left( \dfrac{\dfrac{2y}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}}{\dfrac{{{x}^{2}}+{{y}^{2}}-1}{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}} \right) $ But we are given that amp $ \dfrac{z-1}{z+1} $ = $ \dfrac{\pi }{3} $ . $$\Rightarrow \dfrac{2y}{{{x}^{2}}+{{y}^{2}}-1}=\tan \left( \dfrac{\pi }{3} \right)$$ tan $$\dfrac{\pi }{3}$$ = $ \sqrt{3} $ $$\begin{aligned} & \Rightarrow \dfrac{2y}{{{x}^{2}}+{{y}^{2}}-1}=\sqrt{3} \\\ & \Rightarrow \sqrt{3}{{x}^{2}}+\sqrt{3}{{y}^{2}}-2y-\sqrt{3}=0 \\\ \end{aligned}$$ Thus, by looking at the equation, we can say that z represents a circle. **So, the correct answer is “Option B”.** **Note** : A equation of a straight line is a one-degree equation and equation of circle and equation of pair of straight lines both are 2nd degree equations with x and y both having a second degree. The difference between the equation of a pair of straight lines and the equation of a circle is a term of 2hxy present in the equation of a pair of straight lines. Hence, the equation we got is a circle, as the 2hxy term is absent.