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Question: Show that the perpendicular distance of a point d (d is a complex number) on the Argand plane from t...

Show that the perpendicular distance of a point d (d is a complex number) on the Argand plane from the line az+az+b=0\overline a z + a\overline z + b = 0 (a is a complex number and b is a real) is ad+ad+b2a\dfrac{{\left. {\left| {a\overline d + \overline a d + b} \right.} \right|}}{{2\left| {\left. a \right|} \right.}} .

Explanation

Solution

Hint : In order to solve the question given above, you need to know about complex numbers. They refer to numbers that can be expressed in the form x+iyx + iy , where x and y are real numbers and i is an imaginary unit. You also need to understand the concept of complex plane or argand plane, which refers to a plot of complex numbers as points.

Complete step by step solution:
We know that the equation of the line is: az+az+b=0\overline a z + a\overline z + b = 0 .
Now, we have to put z=x+iyz = x + iy .
The above equation becomes:
a(xiy)+a(x+iy)+b=0a\left( {x - iy} \right) + \overline a \left( {x + iy} \right) + b = 0 .
Therefore, from the above equation, we get that:
(a+a)x+(aa)iy+b=0\left( {a + \overline a } \right)x + \left( {\overline a - a} \right)iy + b = 0 .
Now, let d=d1+id2d = {d_1} + i{d_2} in the equation,
With the help of the above information, you can calculate the distance of the line from d:
(a+a)d1+(aa)id2+b(a+a)2+[(aa)i]2\dfrac{{\left. {\left| {\left( {a + \overline a } \right)} \right.{d_1} + \left( {\overline a - a} \right)i{d_2} + b} \right|}}{{\sqrt {{{\left( {a + \overline a } \right)}^2}} + {{\left[ {\left( {\overline a - a} \right)i} \right] }^2}}} .
This can be written as:
a(d1id2)+a(d1+id2)+b(a+a)2(aa)2\dfrac{{\left. {\left| {a\left( {{d_1} - i{d_2}} \right) + \overline a \left( {{d_1} + i{d_2}} \right)} \right. + b} \right|}}{{\sqrt {{{\left( {a + \overline a } \right)}^2} - {{\left( {\overline a - a} \right)}^2}} }}
ad+ad+b4aa\Rightarrow \dfrac{{\left. {\left| {a\overline d + \overline a d + b} \right.} \right|}}{{\sqrt {4\overline a a} }} .
This gives us:
ad+ad+b4a2\dfrac{{\left. {\left| {a\overline d + \overline a d + b} \right.} \right|}}{{\sqrt {4\left| {{{\left. a \right|}^2}} \right.} }}
ad+ad+b2a\Rightarrow \dfrac{{\left. {\left| {a\overline d + \overline a d + b} \right.} \right|}}{{2\left| {\left. a \right|} \right.}} , which is our required answer.
Hence, proved.

Note : To solve questions similar to the one given above, you need to have basic understanding about a few important topics. These topics are: 1) complex numbers: they are numbers which include both the real and the imaginary parts. They can be expressed in the form x+iyx + iy , where x and y are real numbers and i is an imaginary unit and 2) argand plane: it refers to a diagram which contains the plot of complex numbers as points.