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# The mirror image of the curve $\arg \left( {\dfrac{{z + i}}{{z - 1}}} \right) = \dfrac{\pi }{4}$, $i = \sqrt { - 1}$ in the line $x - y = 0$ isA.$\arg \left( {\dfrac{{z + i}}{{z + 1}}} \right) = \dfrac{\pi }{4}$B.$\arg \left( {\dfrac{{z + 1}}{{z - i}}} \right) = \dfrac{\pi }{4}$C.$\arg \left( {\dfrac{{z - i}}{{z + 1}}} \right) = \dfrac{\pi }{4}$D.$\arg \left( {\dfrac{{z + i}}{{z - 1}}} \right) = \dfrac{\pi }{4}$

Last updated date: 14th Sep 2024
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Hint: First, we will replace $\bar z$ for $z$ in the given equation and then we will use the property of the argument of conjugate of complex numbers, $\arg \left( {\bar z} \right) = - \arg \left( z \right)$. Then we will simplify the obtained expression to find the mirror image.

We are given that the curve $\arg \left( {\dfrac{{z + i}}{{z - 1}}} \right) = \dfrac{\pi }{4}$, $i = \sqrt { - 1}$.
We know that the mirror image of any complex number $z = a + ib$ on the real axis is the conjugate of $z$, i.e., $\overline z = a - ib$.
Now, for determining the mirror image of the curve, we will replace $\bar z$ for $z$ in the given equation, we get
$\Rightarrow \arg \left( {\dfrac{{\bar z + i}}{{\bar z - 1}}} \right) = \dfrac{\pi }{4}$
Using the property of the argument of conjugate of complex numbers, $\arg \left( {\bar z} \right) = - \arg \left( z \right)$ in the above equation, we get
$\Rightarrow \arg \left( {\dfrac{{ - z + i}}{{ - z - 1}}} \right) = \dfrac{\pi }{4} \\ \Rightarrow \arg \left( {\dfrac{{ - \left( {z - i} \right)}}{{ - \left( {z + 1} \right)}}} \right) = \dfrac{\pi }{4} \\ \Rightarrow \arg \left( {\dfrac{{z - i}}{{z + 1}}} \right) = \dfrac{\pi }{4} \\$
Therefore, we have found that the mirror image of the curve is $\arg \left( {\dfrac{{z - i}}{{z + 1}}} \right) = \dfrac{\pi }{4}$.
Note: In such problems, you may get confused amongst the properties used. We can also solve this question using the property of argument of complex numbers like $\arg \left( {\dfrac{{{z_1}}}{{{z_2}}}} \right) = - \arg \left( {\dfrac{{{z_2}}}{{{z_1}}}} \right)$. The mirror image is when we draw a curve, for example the given curve is a circle and it lies in the first quadrant, then its mirror image lies in the 4th quadrant. So the coordinate of $x$–axis will remain the same but the coordinate on the $y$–axis will change its sign.