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Thread: complexe

  1. #1
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    complexe

    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
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  2. #2
    MHF Contributor red_dog's Avatar
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    What is the second number?
    Is $\displaystyle Z^2, \ 2Z, \ Z-2, \ Z+2$ or what? I don't understand.
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  3. #3
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    Quote Originally Posted by red_dog View Post
    What is the second number?
    Is $\displaystyle Z^2$ or what? I don't understand.
    it's $\displaystyle Z^2$
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  4. #4
    MHF Contributor red_dog's Avatar
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    So, we have to find $\displaystyle Z$ such as $\displaystyle |Z|=|Z^2|=|Z+1|$.
    The first equality can be written as $\displaystyle |Z|=|Z|^2\Rightarrow |Z|=0$ or $\displaystyle |Z|=1$.
    If $\displaystyle |Z|=0\Rightarrow Z=0\Rightarrow |Z+1|=1$, but $\displaystyle |Z+1|$ must be 0.

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

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

    So, there are two complex numbers satisfying the problem:
    $\displaystyle Z=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ and $\displaystyle Z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$
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  5. #5
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    ok Many Thanks i was the only one who made it
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