Find the set of all complex numbers $z$ such that, $$1,z,1+z^2$$ are collinear.
Ideas,,
I dont know how start it.
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do you mean all complex numbers $z$ ? – Furrane Mar 19 '17 at 23:17
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yes, I am sorry – mathnw Mar 19 '17 at 23:19
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2What is your level in math ? Are you new to complex numbers ? – Furrane Mar 19 '17 at 23:21
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yes, new.. :( ... – mathnw Mar 19 '17 at 23:23
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Well finding this set might not be very easy then, but surely you can find a part of it, for example what happens if $z=3$ ? Can you generalise ? Then, if you write $z=a+ib$, what is $z^2$ and $1+z^2$ ? From then, how can you check if three points on a graph are colinears ? – Furrane Mar 19 '17 at 23:27
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@Kaster this is not the whole set though, only a part of it =) – Furrane Mar 19 '17 at 23:27
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Literally all of them, because any two points are collinear. – eyeballfrog Mar 19 '17 at 23:29
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Take $z=i$, then the three points are not collinear. – copper.hat Mar 19 '17 at 23:29
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Yes indeed. Carefull though, we need $1,z,1+z^2$ to be colinears, not $z=k\cdot 1$ – Furrane Mar 19 '17 at 23:30
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Ok guys I need to go to sleep. I'll give you a gift : $1+i$ Good night =) – Furrane Mar 19 '17 at 23:31
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@Furrane how's $1+i$ collinear with 1? – Kaster Mar 19 '17 at 23:32
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1@Kaster $\mathbb R^2$ is just as good of a representation of $\mathbb C$ as $\mathbb E^2$. – eyeballfrog Mar 19 '17 at 23:35
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OK: The distinct numbers s $a, b, c$ are collinear in the complex plane if the vector from $a$ to $c$ is a scalar multiple of the vector from $a$ to $b$. Thus ${(c-a)}/{(b-a)}$ is to be a real number, the quotient has to have zero imaginary part. Does this help? – Oscar Lanzi Mar 19 '17 at 23:41
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@eyeballfrog I see. Was thinking about these numbers as vectors, which makes OP's question less meaningful, so I guess you're (and everyone answering) right. – Kaster Mar 20 '17 at 00:00
4 Answers
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Look at the real & imaginary parts of each number $1=(1,0)$, $z=(x,y)$ and $1+z^2=(1+x^2-y^2,2xy)$. If these points are colinear then they will have the same gradient between them so
\begin{eqnarray*}
m= \frac{y}{x-1}= \frac{y(2x-1)}{1+x^2-y^2-x}
\end{eqnarray*}
Firstly note that $y=0$ is a solution & corresponds to $z$ being real & all three numbers lying on the real line.
Now after a little algebra $y^2=x(2-x)$ and which can be written as $(x-1)^2+y^2=1$ So the other solution is a circle of radius one centered at $1$.
Donald Splutterwit
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It is clear that if $z$ is real then the three points lie on a line (the real axis), so we may assume that $z$ is not real.
Then the three points are collinear iff there is some real $t$ such that $1+z^2 = (1-t) 1 + t z$, and we see that this is true iff $t= {z^2 \over z-1}$ and $t$ is real.
We see that ${z^2 \over z-1}$ is real iff $z^2(\bar{z}-1)$ is real.
Letting $z=re^{i\theta}$ we see that $z^2(\bar{z}-1)$ is real iff $r^2 \sin \theta = \sin (2 \theta)$ iff $r = 2 \cos \theta$ and $ r \ge 0$.
Since this gives $z=r e^{i \theta} = 1+\cos(2 \theta)+i\sin(2 \theta)$ we see that the solution is $\mathbb{R} \cup \{ z | |z-1| = 1 \}$.
copper.hat
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we are answering the same question again ! Nice to see you again ... I get lost on the last line of your answer & cannot reconcile it with my answer. – Donald Splutterwit Mar 19 '17 at 23:53
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By shifting the line by $-1$, we see that the question is the same as asking when $$0,z - 1, z^2$$ are collinear. This happens if and only if $z^2$ is a scalar multiple of $z - 1$, i.e. $$z^2 = c(z - 1)$$ for some $c \in \mathbb{R}$. Clearly this holds when $z = 1$. If $z \neq 1$, the statement holds if and only if $$\textrm{Im}\left(\frac{z^2}{z - 1}\right) = 0.$$ You can work this out for a more explicit description if desired.
Ethan Alwaise
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Or, from the second equation then solve by quadratic formula to get $z = \frac{1}{2}(c \pm \sqrt{c^2 - 4c})$. If $c^2 - 4c \ge 0$ this just gives a real number; if $c^2 - 4c < 0$ then $z = \frac{1}{2}(c \pm i \sqrt{4c - c^2}) = x \pm i \sqrt{2x - x^2}$. – Daniel Schepler Mar 19 '17 at 23:49
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A classical condition for 3 complex numbers $z_1,z_2,z_3$ to be aligned (Equation of line in form of determinant) is :
$$\det\begin{pmatrix}z_1&z_2&z_3\\\bar z_1&\bar z_2&\bar z_3\\1&1&1\end{pmatrix}=0$$
Taking $z=re^{i t}$, this condition becomes:
$$\begin{vmatrix}1&re^{i t}&1+r^2 e^{2it}\\1&re^{-i t}&1+r^2 e^{-2it}\\1&1&1\end{vmatrix}=0$$
Which is equivalent, after development, and using Euler formulas, to this equation:
$$-2i r^2 \sin t (r-2 \cos t)=0 $$
which gives rise to 3 types of solutions:
$r=0$ i.e., point $0$.
$\sin t=0 \iff t=k \pi$ : all real $z$, i.e., the $x$ axis.
$r=2 \cos t$ that corresponds to all $z$ belonging to the circle with center in $(1,0)$ with radius 1.
Explanation for the circle: if $z=r e^{it}$ is such that $r=2 \cos t$, we can write: $z=2 \cos t e^{it}=(e^{-it}+e^{it})e^{it}$ (using Euler formula for $\cos t$); thus, expanding : $z=1+e^{2it}$ which is clearly the translated curve of unit circle by vector $(1,0)$.
Jean Marie
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