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1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.

I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but I don't know how to take advantage from $z_1+z_2+z_3=0$

2)Let $z=\cos\alpha+i\sin\alpha$ where $\alpha \in 0,2\pi$ then find $\arg(z^2-z)$

I come to this siutation $\displaystyle z^2-z=-2\sin{\frac{1}{2}x}(\sin{\frac{3}{2}x}+i\cos{\frac{3}{2}x})=-2\sin{\frac{1}{2}x}(\cos(\frac{\pi}{2}-{\frac{3}{2}x})+i\sin({\frac{\pi}{2}-\frac{3}{2}x}))$ so $\displaystyle 0\le\frac{\pi}{2}-\frac{3}{2}x\le2\pi$ so $\displaystyle\frac{\pi}{3}\ge x \ge - \pi$ so $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???

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  • $\begingroup$ Hint for $1$:You know that $e^{i\alpha},e^{i(\alpha+2\pi 1/3)},e^{i(\alpha+2\pi 2/3)}$ satisfy the requirements. Then assume there are other solutions and find a contradiction. $\endgroup$ – flawr Oct 26 '14 at 19:30
  • $\begingroup$ Hint for 2: Write z as $z=e^{i\alpha}$ Then $z^2-z = e^{i2\alpha}-e^{i\alpha}$ $\endgroup$ – flawr Oct 26 '14 at 19:41
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Let: $z_1 =e^{ia} ; z_2 = e^{ib}; z_3 = e^{ic}$

$ z_1 +z_2 = e^{i\frac{a+b}{2}}*(e^{i\frac{a-b}{2}} + e^{-i\frac{(a-b)}{2}}) = e^{i\frac{a+b}{2}}*2*cos(\frac{a-b}{2}) = -z_3 $

=> $|2*cos(\frac{a-b}{2})| = |-z_3| = |z_3| = 1$ ,

If $ cos(\frac{a-b}{2}) =\frac{1}{2} $ -> $a = b \pm \frac{2\pi}{3}$ $mod(2\pi)$

here without loss of generality you can assume a= b+ $\frac{2\pi}{3}$ $ mod(2\pi)$ (the other case is the same)

you get : $\frac{a+b}{2} = c+\pi$ $ mod(2\pi)$ -> b+ $\frac{\pi}{3} = c + \pi$ $ mod(2\pi)$ -> $ b = c + \frac{2\pi}{3} $ $ mod(2\pi)$

You get your equilateral triangle, since you proved that you can rotate of $\frac{2\pi}{3}$ to pass from one point to another. The other cases are exactly the same.

As for 2) , I would use : $z= e^{ia}$

$z^2 - z = e^{2ia} - e^{ia}$ = $e^{\frac{3}{2}ia}*2i*sin(\frac{a}{2}) $ = $ e^{(\frac{3}{2}a + \frac{\pi}{2})i}*2*sin(\frac{a}{2}) $. The sign of the sin is the only thing you have take into account to evaluate correctly the argument. If it is negative, you add $\pi$, else you already have your argument

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  • $\begingroup$ why $|\displaystyle 2\cos(\frac{a-b}{2})|=1$ ?? $\endgroup$ – Mario Oct 26 '14 at 21:57
  • $\begingroup$ You equal the module of both part of the relation : $e^{i\frac{a+b}{2}}*2*cos(\frac{a-b}{2}) = -z_3$ $\endgroup$ – mvggz Oct 26 '14 at 22:46
  • $\begingroup$ I've edited my answer about your second question, if it helps you $\endgroup$ – mvggz Oct 27 '14 at 12:58
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Here my extended hint for $2$:

Notice that $z^2-z = z(z-1) = e^{i\alpha}(e^{i\alpha}-1)$. And since $e^{i(\alpha+\beta)} = e^{i\alpha} e^{i\beta}$ therefore $\arg(ab)=\arg(a)+\arg(b)$

So $\arg(z^2-z) = \alpha + \arg(e^{i\alpha}-1)$

You will now find geometrically that $\arg(e^{i\alpha}-1) = \pi/2 + \alpha/2$ (I hope this is correct.)

You just have to consider the triangle $(0,e^{i\alpha},1)$

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  • $\begingroup$ OK, I'm getting lost when you say that $\arg(e^{i\alpha}-1) = \pi/2 + \alpha/2$ how do you know that ? $\endgroup$ – Mario Oct 26 '14 at 20:01
  • $\begingroup$ Make a drawing of the said triangle, you'll notice that it is an isoscles triangle. If you dissect $\alpha$ you get a triangle with a right angle. In this one you can use the sum of the internal angles. $\endgroup$ – flawr Oct 26 '14 at 20:07
  • $\begingroup$ @flawr If $\alpha=-\pi$ then $Arg(e^{i\alpha}-1)=Arg(-2)=\pi$ and your formula gives $\pi/2 - \pi/2=0$ $\endgroup$ – Harto Saarinen Oct 26 '14 at 20:20
  • $\begingroup$ Ok that is an ambiguity but $\alpha$ is assumed to be nonnegative. $\endgroup$ – flawr Oct 26 '14 at 20:45
  • $\begingroup$ Well same happens if $\alpha=\pi$. $\endgroup$ – Harto Saarinen Oct 26 '14 at 21:32

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