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 Complex numbers and geometry
04-22-2010, 09:50 PM
Post: #1
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
Complex numbers and geometry
(1) Let $$a,b,c \in \mathbf{C}, \quad |a|=|b|=|c|=1$$, show that $$a,b,c$$ represent vertices of a equilateral triangle iff
$$\quad \quad \quad a+b+c=0$$
(2) Find the geometry corresponding to (a) $$I(z^2)>2$$; (b) $$\left|\displaystyle{\frac{z-1}{z+1}}\right|<1$$
(3)A triangle's two vertices are $$a,b$$, find the region for the 3rd vertex $$z$$.
(4) If $\sum_1^6 z_j = 0,\; |z_k| \le 1 \; (\forall k)$ then $|x_i+x_j+x_k|\le 1$ for some distinct $i,j,k \in \{1,2,3,4,5,6\}$
04-23-2010, 09:31 AM
Post: #2
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
RE: Complex numbers and geometry
Let $x,y,z$ be any permutation of $a,b,c$, we have
(1)"If" part: $\overline{x}(x+y+z)=0 \Rightarrow \overline{x}(y+z)=-1=x( \overline{y}+\overline{z})$ So
$\quad |x-y|^{2}-|y-z|^{2}= ((x-y)(\overline{x}-\overline{y})+(x-z)(\overline{x}-\overline{z}))-((y-z)(\overline{y}-\overline{z})+(x-z)(\overline{x}-\overline{z}))$
$\quad =(4-x(\overline{y}+\overline{z})-\overline{x}(y+z))-(4-z(\overline{x}+\overline{y})-\overline{z}(x+y))=6-6=0$
$\quad$thus $\quad \quad |x-y|=|y-z|$
$\quad$If only: Let $\sigma = x+y+z, \quad \rho = |\sigma|$
$\quad |x-y|^2+|x-z|^2=|y-z|^2+|x-z|^2 \Rightarrow x( \overline{y}+\overline{z})+\overline{x}(y+z)=z( \overline{x}+\overline{y})+\overline{z}(x+y)$
$\quad \overset{2x \overline{x}=2z \overline{z}}{\Rightarrow} x \overline{\sigma}+\overline{x}\sigma=z\overline{\sigma}+\overline{z}\sigma \Rightarrow (x-z) \overline{\sigma} = -( \overline{x-z} ) \sigma \Rightarrow ((x-z) / |x-z|)^2 \rho^2 = \sigma^2$
$\quad \quad \Rightarrow \sigma = 0$ since $$x,y,z$$ is an arbitrary permutation of $$a,b,c$$
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