I have an objective type question :-
The set complex numbers $z$ satisfying the equation:-
$$(3+7i)z+(10-2i)\bar{z}+100=0$$
represents:-
A) A straight line
B) A pair of intersecting straight lines
C) A point
D) A pair of distinct parallel straight lines
My approach:-
I have put $z=x+iy$ and solved equation to get :-
$$13x-9y+i(5x-7y)+100=0$$
i.e.
$$y=\frac{13}{9}x+\frac{100}{9}\tag{1.}$$&
$$y=\frac{5}{7}x\tag{2.}$$
So eq. $1$ & $ 2$ are two straight lines which are intersecting each other.So my answer is B)
Now i want to ask that ,is my approach and answer correct? if yes ,then is there any simpler approach? and if no then how would i get the answer?
[Math] represented by the set of complex numbers z satisfying the equation $(3+7i)z+(10-2i)\bar{z}+100=0$
complex numbersgeometryplane-geometry
Related Solutions
The theory of Conic Sections stems from ancient times. It is an example of pure mathematics, which has found applications only many centuries after it
has been developed, e.g. with the laws of planet motion as discovered
by Johannes Keppler. But, interesting as it is, we shall leave aside
history and immediately come to core - or rather
cone - business.
The problem is to intersect a
circular cone with a plane and determine the curves of intersection, like shown
in the above picture. This could be done in the way the old Greek mathematicians
did it. But we prefer to take a path that requires less ingenuity and we shall
employ the means of modern analytical geometry instead. With such an approach,
though, one should be prepared for tedious algebra when working out
the details.
Analysis
A circular cone is characterized by the fact that the angle $\phi$ between the cone axis and its surface is a constant. Let the unit vector $\vec{a}$ be the direction of the cone axis and let $\vec{p}$ point to the top vertex of the cone. An arbitrary point at the surface of the cone is pinpointed by $\vec{r}$. Then the following is an equation of the cone surface: $$ (\vec{a}\cdot\vec{r}-\vec{p}) = |\vec{a}||\vec{r}-\vec{p}|\cos(\phi) $$ Square both sides: $$ (\vec{a}\cdot\vec{r}-\vec{p})^2 = (\vec{a}\cdot\vec{a})(\vec{r}-\vec{p}\cdot\vec{r}-\vec{p})\cos^2(\phi) $$ And work out: $$ (\vec{a}\cdot\vec{r})^2 - 2(\vec{a}\cdot\vec{p})(\vec{a}\cdot\vec{r}) + (\vec{a}\cdot\vec{p})^2 = \cos^2(\phi) \left\{ (\vec{r}\cdot\vec{r}) - 2(\vec{p}\cdot\vec{r}) + (\vec{p}\cdot\vec{p})\right\} $$ The unit vector $\vec{a}$ can be written as: $$ \vec{a} = \left[ \cos(\alpha)\cos(\gamma),\cos(\alpha)\sin(\gamma),\sin(\alpha)\right] $$ Where $\alpha$ is the angle between the cone axis and the XY-plane and $\gamma$ is an angle that indicates how the conic section is rotated in the plane. The vector of the top of the cone can be written in its coordinates as: $$ \vec{p} = (p,q,h) $$ Where $h$ is the height of the cone above the XY plane and $(p,q)$ indicates how the conic section is translated in the plane. Last but not least, the vector pointing to the cone surface is written as: $$ \vec{r} = (x,y,z) $$ Where the intersections with the XY plane are found for $z = 0$. Let's do just that and work out the above: $$ \begin{cases} (\vec{a}\cdot\vec{r}) &=& \cos(\alpha)\cos(\gamma)\,x + \cos(\alpha)\sin(\gamma)\,y \\ (\vec{a}\cdot\vec{p}) &=& \cos(\alpha)\cos(\gamma)\,p + \cos(\alpha)\sin(\gamma)\,q + \sin(\alpha)\,h \\ (\vec{r}\cdot\vec{r}) &=& x^2 + y^2 \\ (\vec{p}\cdot\vec{r}) &=& p\,x + q\,y \\ (\vec{p}\cdot\vec{p}) &=& p^2 + q^2 + h^2 \end{cases} $$ Collecting powers of $x$ and $y$ results in: $$ A\,x^2 + B\,xy + C\,y^2 + D\,x + E\,y + F = 0 $$ Where: $$ \begin{cases} A &=& \cos^2(\phi) - \cos^2(\alpha)\cos^2(\gamma) \\ B &=& - 2 \cos^2(\alpha)\cos(\gamma)\sin(\gamma) \\ C &=& \cos^2(\phi) - \cos^2(\alpha)\sin^2(\gamma) \\ D &=& 2 \left\{ \cos(\alpha)\cos(\gamma) (\vec{a}\cdot\vec{p}) - \cos^2(\phi)\,p \right\} \\ E &=& 2 \left\{ \cos(\alpha)\sin(\gamma) (\vec{a}\cdot\vec{p}) - \cos^2(\phi)\,q \right\} \\ F &=& (\vec{p}\cdot\vec{p}) \cos^2(\phi) - (\vec{a}\cdot\vec{p})^2 \end{cases} $$
Meaning
The first three coefficients of the conic section equation are:
$$
\begin{cases}
A &=& \cos^2(\phi) - \cos^2(\alpha)\cos^2(\gamma) \\
B &=& - 2 \cos^2(\alpha)\cos(\gamma)\sin(\gamma) \\
C &=& \cos^2(\phi) - \cos^2(\alpha)\sin^2(\gamma)
\end{cases}
$$
All kind of conics can still be produced if the angles $\phi$ and $\alpha$ are
limited to sensible values:
$$
\begin{cases}
&&0 < \phi < 90^o \quad \Longrightarrow \quad 0 < \cos(\phi) < 1 \\
&&0 \le \alpha \le 90^o \quad \Longrightarrow \quad 0 \le \cos(\alpha) \le 1
\end{cases}
$$
Generality is not affected by these choices. Moreover it is seen from the picture
below that the form of the conic section is determined
by the angles $\phi$ and $\alpha$ and nothing else. Therefore the ratio of the
two angles will be defined here as the excentricity ($\epsilon$) of the
conic section:
$$
\epsilon = \frac{\cos(\alpha)}{\cos(\phi)}
$$
The following relationships exist between the excentricity and the form of a
conic section, as is clear from the picture:
$$
\begin{cases}
\mbox{Circle : }& \alpha = 90^o &\quad \Longleftrightarrow \quad \epsilon = 0 \\
\mbox{Ellipse : }& \alpha > \phi &\quad \Longleftrightarrow \quad \epsilon < 1 \\
\mbox{Parabola : }& \alpha = \phi &\quad \Longleftrightarrow \quad \epsilon = 1 \\
\mbox{Hyperbola : }& \alpha < \phi &\quad \Longleftrightarrow \quad \epsilon > 1
\end{cases}
$$
So far so good. The coefficients $(A,B,C)$ can be combined into
some interesting quantities which are only dependent upon form,
that is: the angles $\phi$ and $\alpha$.
It is remarked in the first place that $(A,B,C)$ are independent of the vector
$\vec{p} = (p,q,h)$ and thus independent of translation and scaling. If we seek
to eliminate any dependence upon the angle of rotation $\gamma$, then we find:
$$
A + C = 2 \cos^2(\phi) - \cos^2(\alpha) = \cos^2(\phi) (2 - \epsilon^2)
$$
This quantity is known (for some good reasons) as the trace of the conic
section. Instead of eliminating the angle of rotation, we could also try
to calculate it.
$$
A - C = - \cos^2(\alpha)\left[\cos^2(\gamma)-\sin^2(\gamma)\right]
= - \cos^2(\alpha)\cos(2\gamma)
$$
There is a striking resemblance with:
$$
B = - \cos^2(\alpha)\sin(2\gamma)
$$
We thus find:
$$
\frac{B}{A - C} = \frac{\sin(2\gamma)}{\cos(2\gamma)} \quad \Longrightarrow \quad
\tan{2\gamma} = \frac{B}{A - C}
$$
Herewith - in principle - the angle of rotation $\gamma$ can be reconstructed
from the conic section equation; provided that $A \neq C$.
Let's proceed with
another quantity that is independent of any rotation.
$$
B^2 - 4 A C = \left[ - 2 \cos^2(\alpha)\cos(\gamma)\sin(\gamma) \right]^2
$$ $$
- 4 \left[ \cos^2(\phi) - \cos^2(\alpha)\cos^2(\gamma) \right]
\left[ \cos^2(\phi) - \cos^2(\alpha)\sin^2(\gamma) \right]
$$
This quantity is known (also for some good reasons) as the determinant or
discriminant of the conic section. Work out:
$$
= 4 \cos^4(\alpha)\cos^2(\gamma)\sin^2(\gamma) - 4 \cos^4(\phi)
- 4 \cos^4(\alpha)\cos^2(\gamma)\sin^2(\gamma) \\
+ 4 \cos^2(\phi)\cos^2(\alpha)\left[\cos^2(\gamma)+\sin^2(\gamma)\right] \\
= - 4\cos^4(\phi) + 4\cos^2(\phi)\cos^2(\alpha) \\
\quad \Longrightarrow \quad B^2 - 4 A C
= 4\cos^2(\phi)\left[\cos^2(\alpha) - \cos^2(\phi)\right]
= \left[2\cos^2(\phi)\right]^2(\epsilon^2 - 1)
$$
The following relationships exist between the discriminant and the form of a
conic section, as is clear from the above:
$$
\begin{cases}
&\mbox{Ellipse}& \quad \Longleftrightarrow \quad \epsilon < 1 \quad \Longleftrightarrow \quad (B^2 - 4 A C) < 0 \\
&\mbox{Parabola}& \quad \Longleftrightarrow \quad \epsilon = 1 \quad \Longleftrightarrow \quad (B^2 - 4 A C) = 0 \\
&\mbox{Hyperbola}& \quad \Longleftrightarrow \quad \epsilon > 1 \quad \Longleftrightarrow \quad (B^2 - 4 A C) > 0
\end{cases}
$$
BONUS on excentricity. $$ (A + C)^2 = 4\cos^2(\phi)\left[\cos^2(\phi) - \cos^2(\alpha)\right] + \cos^4(\alpha) $$ Upon addition this gives: $$ (B^2 - 4 A C) + (A + C)^2 = \cos^4(\alpha) \quad \Longrightarrow \\ \cos(\alpha) = \sqrt{\sqrt{B^2 + (A-C)^2}} $$ About the angle $\phi$ between the cone axis and its surface: $$ A + C = 2\cos^2(\phi) - \sqrt{B^2 + (A-C)^2} \quad \Longrightarrow \quad \cos(\phi) = \sqrt{\frac{(A+C) + \sqrt{B^2 + (A-C)^2}}{2}} $$ Herewith the excentricity $\epsilon$ can be expressed into the coefficients of the conic section equation $(A,B,C)$: $$ \epsilon = \sqrt{\frac{2\sqrt{B^2 + (A-C)^2}} {(A+C) + \sqrt{B^2 + (A-C)^2}}} $$ Many other expressions can be derived, especially for the coefficients $D,E,F$. But I think that deriving them here will disturb a good balance between interesting and cumbersome.
EDIT. Explanation of
$$
\vec{a} = \left[
\cos(\alpha)\cos(\gamma),\cos(\alpha)\sin(\gamma),\sin(\alpha)\right]
$$
See picture. Overline denotes " length of " in:
$$
\overline{OZ}=\overline{OA}\, \sin(\alpha) \quad ; \quad
\overline{OD}=\overline{OA}\, \cos(\alpha) \\
\overline{OX}=\overline{OD}\, \cos(\gamma) \quad ; \quad
\overline{OY}=\overline{OD}\, \sin(\gamma)
$$
With $\overline{OA}=1$ .
one of the tangents from $(3,2)$ touch the circle at $(0,2).$ if you look at the right angle triangle formed by the points $(0,0), (3,2)$ and $(0,2),$ then half the angle between the tangents is $\arctan(2/3).$ now use the formula $\tan(2t) = \frac{2\tan t}{1- \tan^2 t} = \frac{4/3}{1-4/9} = \frac{12}5.$
Best Answer
Once you get to the step where you have the equations of the two lines, there is still more work to do. Or there would be if you had to actually find the point of intersection. But the answer is C, not B. Everything you've done up to that point is correct and I'm not sure if there's a better way to do it.
So why is the answer C and not B? You started with the given equation: $$(3+7i)z + (10-2i)\overline{z} + 100 = 0$$ Then you let $z = x+iy$: $$(3+7i)(x+iy) + (10-2i)(x-iy) + 100 = 0$$ Then you did some algebra and ended up with: $$13x - 9y + i(5x-7y) + 100 = 0 \tag{*}$$ Then you equated real and imaginary parts to obtain two equations, and the key here is that both equations must be true at the same time in order for equation (*) above to hold. And this only happens at the point of intersection.