complexe

**iceman1** · September 24th 2007, 06:00 AM

Hello Guys

Hope you are Fine

How to determine the Numbers complexes Z such as:
Z, Z (2), 1+Z has the same Module?

Thanks in Advance Guys

**red_dog** · September 24th 2007, 08:30 AM

What is the second number?
Is $Z^2, \ 2Z, \ Z-2, \ Z+2$ or what? I don't understand.

**iceman1** · September 24th 2007, 09:40 AM

Originally Posted by red_dog

What is the second number?
Is $Z^2$ or what? I don't understand.

it's $Z^2$

**red_dog** · September 24th 2007, 10:06 AM

So, we have to find $Z$ such as $|Z|=|Z^2|=|Z+1|$ .
The first equality can be written as $|Z|=|Z|^2\Rightarrow |Z|=0$ or $|Z|=1$ .
If $|Z|=0\Rightarrow Z=0\Rightarrow |Z+1|=1$ , but $|Z+1|$ must be 0.

If $|Z|=1=|Z+1|$ and $Z=x+yi$ , we have to solve the system:
$\displaystyle\left\{\begin{array}{ll}\sqrt{x^2+y^2 }=1\\\sqrt{(x+1)^2+y^2}=1\end{array}\right.$ or

$\displaystyle\left\{\begin{array}{ll}x^2+y^2=1\\x^ 2+y^2+2x=0\end{array}\right.$
Substracting the first equation from the second we get $x=-\frac{1}{2}$ .
Now, plug $x$ in the first equation and solve for $y$ .
We get $y=-\frac{\sqrt{3}}{2}$ or $y=\frac{\sqrt{3}}{2}$ .

So, there are two complex numbers satisfying the problem:
$Z=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ and $Z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$

**iceman1** · September 26th 2007, 05:51 AM

ok Many Thanks

i was the only one who made it

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