Question

The mirror image of the curve \[\arg \left( {\dfrac{{z + i}}{{z - 1}}} \right) = \dfrac{\pi }{4}\], \[i = \sqrt { - 1} \] in the line \[x - y = 0\] is
A.\[\arg \left( {\dfrac{{z + i}}{{z + 1}}} \right) = \dfrac{\pi }{4}\]
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.\[\arg \left( {\dfrac{{z + i}}{{z - 1}}} \right) = \dfrac{\pi }{4}\]

Answer 1

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.

Complete step-by-step answer:
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 \[\arg \left( {\dfrac{{z - i}}{{z + 1}}} \right) = \dfrac{\pi }{4}\].

Hence, option C is correct.

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.