Complexe

**iceman1** · September 29th 2007, 03:19 AM

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

**topsquark** · September 29th 2007, 04:15 AM

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
$\frac{1 - iz}{1 + iz} = 1$

$1 - iz = 1 + iz$

Thus
$-z = z$

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

-Dan

**red_dog** · September 29th 2007, 07:03 AM

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

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

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

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

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