Thread: Complexe

sorry Guys but my math teachers gave 30 problems For us he thinks that i don't have only Math to do it , i don't have a lot of times i'll wake up today to study lol.

anyway i stopped in 3 question - Hard For me

z is a complex number, prove if
{1 - iz
-------} = 1, then z is real
1 + iz


my second question:
that is to say A and B are two points of the plan of respective affixes zA, zB
will
a.Prove that [zA-ZB] = AB
B. find all the numbers complexes z such as the triangle whose tops are the images of 1, Z, z^2 is equilateral


and my last question:

Z and z' are two complex numbers unspecified.
Show that [z+z'] ^2 + {z-z'] ^2 = 2 {[Z} 2 + {z'} ^2}


P.s: i don't know how to write them so [ ] =




Many Thanks Guys

Originally Posted by iceman1

z is a complex number, prove if
{1 - iz
-------} = 1, then z is real
1 + iz

I'll do better than that:
If
$\displaystyle \frac{1 - iz}{1 + iz} = 1$

$\displaystyle 1 - iz = 1 + iz$

Thus
$\displaystyle -z = z$

Thus $\displaystyle z = 0$. Yeah, it's a real number!

-Dan

1). I think it's about $\displaystyle \displaystyle\left|\frac{1-iz}{1+iz}\right|=1$.
Then $\displaystyle |1-iz|=|1+iz|$. (1)
Let $\displaystyle z=x+yi\Rightarrow 1-iz=(1+y)-ix, \ 1+iz=(1-y)+ix$.
Then the equality (1) becomes $\displaystyle \sqrt{(1+y)^2+x^2}=\sqrt{(1-y)^2+x^2}$.
Square both sides: $\displaystyle 1+2y+y^2+x^2=1-2y+y^2+x^2\Rightarrow 4y=0\Rightarrow y=0$.
So, $\displaystyle z=x\in\mathbf{R}$.

2)
a) Let $\displaystyle A(x_A,y_B),B(x_B,y_B)$
Then $\displaystyle z_A=x_A+y_Ai, \ z_B=x_B+y_Bi$.
$\displaystyle |z_A-z_B|=|x_A-x_B+(y_A-y_B)i|=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}=AB$.

b) Let $\displaystyle A(1),B(z),C(z^2)$
Then ABC is equilateral$\displaystyle \Leftrightarrow AB=AC=BC\Leftrightarrow |z-1|=|z^2-1|=|z^2-z|$ (2)
The equalities (2) can be written as $\displaystyle |z-1|=|z-1|\cdot |z+1|=|z|\cdot |z-1|$.
Suppose that $\displaystyle z\neq 1$, otherwise the triangle degenerates to a point. Then $\displaystyle |z-1|\neq 0$, so we can divide all three members by $\displaystyle |z-1|$.
We get
$\displaystyle 1=|z+1|=|z|$.
Let $\displaystyle z=x+yi$.
Then $\displaystyle \left\{\begin{array}{ll}(x^2+y^2=1\\(x+1)^2+y^2=1\ end{array}\right.$.
Solving the system, we get $\displaystyle \displaystyle x=-\frac{1}{2}, \ y=\pm\frac{\sqrt{3}}{2}$.

3) We'll use the relation $\displaystyle z\cdot\overline{z}=|z|^2$.
$\displaystyle |z+z'|^2+|z-z'|^2=(z+z')(\overline{z+z'})+(z-z')(\overline{z-z'})=(z+z')(\overline{z}+\overline{z'})+(z-z')(\overline{z}-\overline{z'})=$
$\displaystyle =z\cdot\overline{z}+z\cdot\overline{z'}+z'\cdot\ov erline{z}+z'\cdot\overline{z'}+z\cdot{z}-z\cdot\overline{z'}-z'\cdot\overline{z}+z'\cdot\overline{z'}=$
$\displaystyle =2(z\cdot\overline{z}+z'\cdot\overline{z'})=2(|z|^ 2+|z'|^2)$