Question

Find the center of the arc represented by $\arg \dfrac{{(z - 3i)}}{{(z - 2i + 4)}} = \dfrac{\pi }{4}$
(A) $9i - 5$
(B)$\dfrac{1}{2}(9i - 5)$
(C)$- 9i + 5$
(D)$\dfrac{1}{2}( - 9i + 5)$

Answer 1

Substitute $z = x + iy$ and simplify the fraction. Use the fact if $z = a + ib$ where $a \ne 0$ then$\arg (z) = \arg (a + ib) = {\tan ^{ - 1}}\dfrac{b}{a} = \theta$ to obtain the general equation of the circle on which the given arc lies. The required answer is the center of this circle.

Complete step by step answer:
Given that the arc is represented by $\arg \dfrac{{(z - 3i)}}{{(z - 2i + 4)}} = \dfrac{\pi }{4}$ and we need to find the center of this arc.
Now, the complex number z can be expressed as $z = x + iy$ where x and y are real numbers and are called the real and imaginary parts of z respectively.
Substituting the value$z = x + iy$ in the given expression, we get the following equation:
$\dfrac{{(z - 3i)}}{{(z - 2i + 4)}} = \dfrac{{(x + iy - 3i)}}{{(x + iy - 2i + 4)}}$
We will group the real and imaginary parts in the numerator and the denominator using brackets.
$\dfrac{{(z - 3i)}}{{(z - 2i + 4)}} = \dfrac{{[x + i(y - 3)]}}{{[(x + 4) + i(y - 2)]}}$
Then we simplify the fraction further and write it in the form of a + bi.
We can do this by multiplying and dividing with the complex conjugate of the denominator.
The denominator is $(x + 4) + i(y - 2)$.
Therefore, the complex conjugate would be $(x + 4) - i(y - 2)$.
Thus, we get the following step:
$\dfrac{{(z - 3i)}}{{(z - 2i + 4)}} = \dfrac{{[x + i(y - 3)]}}{{[(x + 4) + i(y - 2)]}} = \dfrac{{[x + i(y - 3)]}}{{[(x + 4) + i(y - 2)]}} \times \dfrac{{[(x + 4) - i(y - 2)]}}{{[(x + 4) - i(y - 2)]}}.......(i)$
We will be multiplying the terms in the numerator followed by those in the denominator.
The formula for finding the product of two complex numbers ${z_1} = a + ib$ and ${z_2} = c + id$ is given by ${z_1}{z_2} = (a + ib)(c + id) = (ac - bd) + i(ad + bc).......(1)$
Consider the numerator $[x + i(y - 3)] \times [(x + 4) - i(y - 2)]$
We need to express the numerator in the form of $(a + ib)(c + id)$

$[x + i(y - 3)] \times [(x + 4) - i(y - 2)]$

$= [x + i(y - 3)] \times [(x + 4) + i( - 1)(y - 2)]$

$= [x + i(y - 3)] \times [(x + 4) + i( - y + 2)]$

Now, apply the product formula (1) for complex numbers

$[x + i(y - 3)] \times [(x + 4) - i(y - 2)]$

$= [x(x + 4) - (y - 3)( - y + 2)] + i[x( - y + 2) + (y - 3)(x + 4)]$

Upon further simplification using the identity $(x + a)(x + b) = {x_2} + (a + b)x + ab$, we get:

$[x + i(y - 3)] \times [(x + 4) - i(y - 2)]$

$= [x(x + 4) - (y - 3)( - y + 2)] + i[x( - y + 2) + (y - 3)(x + 4)]$

$= [{x^2} + 4x - (( - {y^2}) + 2y + 3y - 6)] + i[ - xy + 2x + (xy + 4y - 3x - 12)]$

$= [{x^2} + 4x + {y^2} - 2y - 3y + 6] + i[ - xy + 2x + xy + 4y - 3x - 12]$

$= [{x^2} + {y^2} + 4x - 5y - 6] + i[ - x + 4y - 12]$

Thus, we have numerator $= [{x^2} + {y^2} + 4x - 5y - 6] + i[ - x + 4y - 12]$
The formula for finding the product of a complex number $z = a + ib$ and its conjugate $\bar z = a - ib$ is given by $z\bar z = (a + ib)(a - ib) = ({a^2} + {b^2})........(2)$
Consider the denominator $[(x + 4) + i(y - 2)] \times [(x + 4) - i(y - 2)]$
After applying the identity given in (2), we get

$[(x + 4) + i(y - 2)] \times [(x + 4) - i(y - 2)]$

$= {(x + 4)^2} + {(y - 2)^2}$
Thus, equation (i) becomes,
$\dfrac{{(z - 3i)}}{{(z - 2i + 4)}} = \dfrac{{[{x^2} + {y^2} + 4x - 5y - 6] + i[ - x + 4y - 12]}}{{{{(x + 4)}^2} + {{(y - 2)}^2}}} = \dfrac{{{x^2} + {y^2} + 4x - 5y - 6}}{{{{(x + 4)}^2} + {{(y - 2)}^2}}} + i\dfrac{{ - x + 4y - 12}}{{{{(x + 4)}^2} + {{(y - 2)}^2}}} = p + iq....(3)$
We know that if $z = a + ib$ where $a \ne 0$ then$\arg (z) = \arg (a + ib) = {\tan ^{ - 1}}\dfrac{b}{a} = \theta$
Clearly, we can see that $p \ne 0$ in (3) as the denominator of $p$ consists of ${(x + 4)^2}$ which is a non-negative term.
Therefore, using the fact

$\tan \dfrac{\pi }{4} = 1$

we get

$\arg \dfrac{{(z - 3i)}}{{(z - 2i + 4)}} = \dfrac{\pi }{4}$
$\Rightarrow {\tan ^{ - 1}}(p + iq) = \dfrac{\pi }{4}$

$\Rightarrow \dfrac{q}{p} = \tan \dfrac{\pi }{4} = 1$
$\Rightarrow q = p$
$\Rightarrow \dfrac{{ - x + 4y - 12}}{{{{(x + 4)}^2} + {{(y - 2)}^2}}} = \dfrac{{{x^2} + {y^2} + 4x - 5y - 6}}{{{{(x + 4)}^2} + {{(y - 2)}^2}}}$
$\Rightarrow - x + 4y - 12 = {x^2} + {y^2} + 4x - 5y - 6$
$\Rightarrow {x^2} + {y^2} + 5x - 9y + 6 = 0.....(4)$
Now, equation (4) is an equation of a circle.
The required answer is the center of this circle.
The general equation of a circle is $\;{x^2} + {y^2} + 2gx + 2fy + c = 0$ where $\left( { - g, - f} \right)$ represents the coordinates of the center of the circle.
On comparing equation (4) with the general equation, we get $2g = 5 \Rightarrow g = \dfrac{5}{2}$ and $2f = - 9 \Rightarrow f = \dfrac{{ - 9}}{2}$
Therefore, $\left( { - g, - f} \right) = (\dfrac{{ - 5}}{2},\dfrac{9}{2})$

Note: While using the fact if $z = a + ib$ where $a \ne 0$ then $\arg (z) = \arg (a + ib) = {\tan ^{ - 1}}\dfrac{b}{a} = \theta$ one needs to be careful in questions where $a = 0$, i.e. the complex number has no real part. In such cases, this method does not work.