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Thread: please help me

  1. #1
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    please help me

    show that :
    if Re (z1) >0 and Re( z2) >0 then :
    $\displaystyle Arg (z1z2) = $
    $\displaystyle Arg (z1)+ Arg (z2) $
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  2. #2
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    Fact: If $\displaystyle z=r\mathrm e^{\mathrm i\theta}$ where $\displaystyle r>0$ and $\displaystyle -\pi<\theta\leq\pi$ then $\displaystyle r=|z|$ and $\displaystyle \theta=\mathop{arg} z$.

    If $\displaystyle \mathop{arg} z_1=\theta$ and $\displaystyle \mathop{arg} z_2=\phi$ then $\displaystyle z_1=|z_1|\mathrm e^{\mathrm i\theta}$ and $\displaystyle z_2=|z_2|\mathrm e^{\mathrm i\phi}$.

    Therefore $\displaystyle z_1z_2=|z_1||z_2|\mathrm e^{\mathrm i\theta}\mathrm e^{\mathrm i\phi}=|z_1z_2|\mathrm e^{\mathrm i(\theta+\phi)}$.

    If $\displaystyle \mathop{Re}(z_1)>0$ then $\displaystyle -{\textstyle\frac\pi2}<\theta<{\textstyle\frac\pi2}$, and if $\displaystyle \mathop{Re}(z_2)>0$ then $\displaystyle -{\textstyle\frac\pi2}<\phi<{\textstyle\frac\pi2}$. Therefore $\displaystyle -\pi<\theta+\phi<\pi$.

    Thus $\displaystyle \mathop{arg}(z_1z_2)=\theta+\phi=\mathop{arg} z_1+\mathop{arg} z_2$.
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  3. #3
    Super Member malaygoel's Avatar
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    Let $\displaystyle z_1=a_1+ib_1$

    and $\displaystyle z_2=a_2+ib_2$

    be two complex numbers where $\displaystyle a_1,a_2>0$

    $\displaystyle arg(z_1)=tan^{-1}\frac{b_1}{a_1}$

    $\displaystyle arg(z_2)=tan^{-1}\frac{b_2}{a_2}$

    Now,$\displaystyle z_1z_2=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1)$

    $\displaystyle arg(z_1z_2)=tan^{-1}\frac{a_1b_2+a_2b_1}{a_1a_2-b_1b_2}$

    $\displaystyle arg(z_1z_2)=tan^{-1}\frac{b_1}{a_1}+tan^{-1}\frac{b_2}{a_2}$

    I don't know how the property $\displaystyle a_1,a_2>0$ is used.
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  4. #4
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    i Noted that $\displaystyle Arg({Z\scriptstyle1} {Z\scriptstyle2}) \leq Arg({Z\scriptstyle1 }) +Arg({Z\scriptstyle2} ) $
    this true or not ??
    i take some examples like :
    $\displaystyle { Z\scriptstyle1} = -5$
    $\displaystyle { Z\scriptstyle2} = -1-i$ ,
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  5. #5
    Super Member malaygoel's Avatar
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    $\displaystyle z=a+ib=|z|e^{i\theta}$

    There are four cases:
    (1)$\displaystyle a\geq 0,b\geq 0$,
    then we have,$\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$

    (2)$\displaystyle a\leq 0,b\geq 0$
    then we have,$\displaystyle \frac{\pi}{2} \leq \theta \leq {\pi}$

    (3)$\displaystyle a\leq 0,b\leq 0$
    then we have,$\displaystyle \pi \leq \theta \leq \frac{3\pi}{2}$

    (4)$\displaystyle a\geq 0,b\leq 0$
    then we have,$\displaystyle \frac{3\pi}{2} \leq \theta \leq 2\pi$

    Now,
    if sum of arguments($\displaystyle \theta_1,\theta_2$) is less than $\displaystyle 2\pi$,
    then
    $\displaystyle
    Arg({Z\scriptstyle1} {Z\scriptstyle2}) = Arg({Z\scriptstyle1 }) +Arg({Z\scriptstyle2} )
    $

    And,
    if sum of arguments($\displaystyle \theta_1,\theta_2$) is equal to or more than $\displaystyle 2\pi$,
    then
    $\displaystyle
    Arg({Z\scriptstyle1} {Z\scriptstyle2}) = Arg({Z\scriptstyle1 }) +Arg({Z\scriptstyle2} )-2\pi
    $
    $\displaystyle
    Arg({Z\scriptstyle1} {Z\scriptstyle2}) < Arg({Z\scriptstyle1 }) +Arg({Z\scriptstyle2} )
    $

    Quote Originally Posted by flower3
    $\displaystyle
    { Z\scriptstyle1} = -5
    $
    $\displaystyle
    { Z\scriptstyle2} = -1-i
    $
    in the example you provided,
    $\displaystyle \theta_1=\pi$
    $\displaystyle \theta_2=\frac{5\pi}{4}$
    $\displaystyle arg(Z_1Z_2)=\pi +\frac{5\pi}{4}-2\pi=\frac{\pi}{4}$
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  6. #6
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    (3)
    then we have,

    (4)
    then we have,
    these cases are not true because :
    $\displaystyle Arg Z=\theta and -\pi \prec \theta \leq \pi $
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  7. #7
    Super Member malaygoel's Avatar
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    Quote Originally Posted by flower3 View Post
    these cases are not true because :
    $\displaystyle Arg Z=\theta and -\pi \prec \theta \leq \pi $
    there is no condtion like this.

    it varies from 0 to $\displaystyle 2\pi$

    Are you familiar with Argand Plane?

    Then plot -1-i on it and find $\displaystyle \theta$.

    EDIT: you and me are saying the same thing...don't know which is standard.
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