Show that the equation $z=a+2bt+ct^2$ represents a parabola where $a,b,c\in\mathbb{C}$.
My approach: let $a=m+m'i, b=l+l'i, c=n+n'i, z=x+yi$.
Equating real and imaginary parts on both sides, we get
$$x=m+2lt+nt^2 ,y=m'+2n't+n't^2$$ since these are quadratic equations in $t, i$ have compared both the equations and getting straight line as locus but i have to show parabola. please help
Best Answer
Translate the origin of the complex plane by $a$ and define $s\equiv z-a$, your new complex variable in your new coordinate system that is just translation of the old one by $a$:
$z-a\equiv s=2 b t+c t^2$
Let $\Re(s)=x$ and $\Im(s)=y$. Thus, with obvious notation,
$x=2b_x t+c_x t^2$
$y=2b_y t +c_y t^2$
Your goal is now to rigidly rotate the $s$-plane so that $x$ becomes function of $t$ alone, without $t^2$ in it. Then $x$ itself is variable $t$, just like in Cartesian system.
Rotation is done by multiplying by $e^{i\phi}$ with some fixed constant $\phi$. Proof: $se^{i\phi}=re^{i\alpha}e^{i\phi}=re^{i(\alpha+\phi)}$. So everything is rotated by angle $\phi$ about the origin.
So multiply $s=x+iy$ by some $\cos\phi+i\sin\phi$, the result reads $s'=x\cos\phi-y\sin\phi + i (y\cos\phi+x\sin\phi)$. Hence,
$x'= 2b_x t\cos\phi+c_x t^2\cos\phi-2b_y t\sin\phi-c_y t^2\sin\phi$
$y'=2b_y t\cos\phi+c_y t^2\cos\phi+2b_x t \sin\phi+c_x t^2\sin\phi$
Terms with $t^2$ disappear from $x$ if you choose to use angle of rotation $\phi$ such that
$c_x\cos\phi=c_y\sin\phi$
or equivalently
$c_y=c_x \cot\phi$
Do notice that $\cot\phi=c_y/c_x$ and so is neither $0$ nor $\infty$ if $c_x\ne 0$ and $c_y\ne 0$. Hence both sine and cosine do not vanish.
Then
$x'= 2t(b_x \cos\phi-2b_y \sin\phi)$
$y'=2b_yt\cos\phi+c_x t^2 \cos^2\phi/\sin\phi+2b_x t \sin\phi+c_x t^2\sin\phi=\\ 2t(b_y\cos\phi+b_x \sin\phi)+c_x t^2 \sin\phi$
Hence $y'$ is quadratic in $t$ whilst $x'$ is only linear in $t$. It's parabola alright.