Question

If $z$ is a complex number, then the locus of the point $z$ satisfying $\arg \left(\frac{z-i}{z+i}\right)=\frac{\mathrm{\pi }}{4}$ is a

• A. circle with centre $\left(-1,0\right)$ and radius $\sqrt{2}$
• B. circle with centre $\left(0,0\right)$ and radius $\sqrt{2}$
• C. circle with centre $\left(0,1\right)$ and radius $\sqrt{2}$
• D. circle with centre $\left(1,1\right)$ and radius $\sqrt{2}$

Answer 1

The correct option is A

circle with centre $\left(-1,0\right)$ and radius $\sqrt{2}$

Explanation for the correct option.

Step 1. Simplify the given equation using the properties of argument.

Let $z=x+iy$ be a complex number.

Now, using $a\mathrm{rg}\left(\frac{z_1}{z_2}\right)=\arg \left(z_1\right)-\arg \left(z_2\right)$, the given equation $\arg \left(\frac{z-i}{z+i}\right)=\frac{\mathrm{\pi }}{4}$ can be written as:

$$\begin{gathered} \arg \left(\frac{z-i}{z+i}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \arg \left(z-i\right)-\arg \left(z+i\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \arg \left(x+iy-i\right)-\arg \left(x+iy+i\right)=\frac{\mathrm{\pi }}{4};\left[z=x+iy\right] \\ \Rightarrow \arg \left(x+\left(y-1\right)i\right)-\arg \left(x+\left(y+1\right)i\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{y-1}{x}\right)-{\tan }^{-1}\left(\frac{y+1}{x}\right)=\frac{\mathrm{\pi }}{4};\left[\arg \left(x+iy\right)={\tan }^{-1}\left(\frac{y}{x}\right)\right] \end{gathered}$$

Step 2. Use the property of the inverse tangent function and form the equation.

Now using the relation: ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$ the equation can be simplified as:

$$\begin{gathered} {\tan }^{-1}\left(\frac{y-1}{x}\right)-{\tan }^{-1}\left(\frac{y+1}{x}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{y-1}{x}}-{\displaystyle \frac{y+1}{x}}}{1+{\displaystyle \frac{y-1}{x}}·{\displaystyle \frac{y+1}{x}}}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{{\displaystyle \frac{y-1-y-1}{x}}}{{\displaystyle \frac{x^2+\left(y-1\right)\left(y+1\right)}{x^2}}}=\tan \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \frac{x\left(-2\right)}{x^2+y^2-1}=1 \\ \Rightarrow -2x=x^2+y^2-1 \\ \Rightarrow 0=x^2+2x+y^2-1 \end{gathered}$$

Step 3. Write the equation in standard form and find the locus of the point $z$.

Now write the equation $x^2+2x+y^2-1=0$ in standard form.

$\begin{array}{rcl}{x}^2+2x+y^2-1& =& 0\\ & \Rightarrow & \left(x^2+2x+1\right)+y^2-1-1=0\\ & \Rightarrow & {\left(x+1\right)}^2+y^2-2=0\\ & \Rightarrow & {\left(x+1\right)}^2+y^2={\left(\sqrt{2}\right)}^2\end{array}$

Now comparing the equation ${\left(x+1\right)}^2+y^2={\left(\sqrt{2}\right)}^2$ with the standard equation of circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ it is found that the centre of the circle is $\left(-1,0\right)$ and its radius is $\sqrt{2}$.

The locus of the point $z$ is a circle with centre $\left(-1,0\right)$ and radius $\sqrt{2}$ and is given by the equation $x^2+2x+y^2-1=0$.

Hence, the correct option is A.