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Thread: Complex Numbers - Find z & w

  1. #1
    Newbie Flamuri's Avatar
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    Red face Complex Numbers - Find Z & W

    Assign Z & W [ Complex numbers ] if

    $\displaystyle z + w = 1 - i$ ;

    $\displaystyle argZ = \frac{\pi}{6} $ ;

    $\displaystyle argW = \frac{5\pi }{3}$

    Have tried so many times, but no results, can someone please help me explaining those kind of tasks

    Thanks,
    Flamuri
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  2. #2
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    Re: Complex Numbers - Find Z & W

    Well, we can see that $\displaystyle \displaystyle z = r_1\left(\cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}}\right) = r_1\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \frac{\sqrt{3}}{2}r_1 + \frac{1}{2}ir_1$ and $\displaystyle \displaystyle w = r_2\left(\cos{\frac{5\pi}{3}} + i\sin{\frac{5\pi}{3}}\right) = r_2\left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = \frac{1}{2}r_2 - \frac{\sqrt{3}}{2}ir_2$.

    So $\displaystyle \displaystyle z + w = \frac{\sqrt{3}}{2}r_1 + \frac{1}{2}r_2 + \left(\frac{1}{2}r_1 - \frac{\sqrt{3}}{2}r_2\right)i$.

    Now equate your real and imaginary parts to give you two equations to solve for $\displaystyle \displaystyle r_1$ and $\displaystyle \displaystyle r_2$ and you will have $\displaystyle \displaystyle z$ and $\displaystyle \displaystyle w$.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Complex Numbers - Find Z & W

    modified.
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  4. #4
    Newbie Flamuri's Avatar
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    Re: Complex Numbers - Find Z & W

    Wow, I've used a completely different way [for me a bit harder way] from this that you have used, but finally found a result:

    If

    $\displaystyle z = a + bi$
    $\displaystyle w = x + yi$

    So:

    $\displaystyle z = - \frac{\sqrt{3}+3}{2} + \frac{\sqrt{3}-1}{2}i$

    $\displaystyle w = \frac{\sqrt{3}+3}{2} + \frac{\sqrt{3}+1}{2}i$

    Don't know how to prove if i'm right, maybe you can help me again,
    Thanks.
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    Re: Complex Numbers - Find Z & W

    Quote Originally Posted by Flamuri View Post
    Assign Z & W [ Complex numbers ] if
    $\displaystyle z + w = 1 - i$ ;

    $\displaystyle argZ = \frac{\pi}{6} $ ; $\displaystyle argW = \frac{5\pi }{3}$
    I would do it this way. Let $\displaystyle z=a+bi~\&~w=c+di$.

    From the given $\displaystyle \begin{align*} a+c&= 1 \\ b+d &=-1 \end{align*}$ and $\displaystyle \frac{b}{a} = \tan \left( {\frac{\pi }{6}} \right)\;\& \,\frac{d}{c} = \tan \left( {\frac{{5\pi }}{3}} \right)$
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    Re: Complex Numbers - Find Z & W

    Quote Originally Posted by Flamuri View Post
    Wow, I've used a completely different way [for me a bit harder way] from this that you have used, but finally found a result:

    If

    $\displaystyle z = a + bi$
    $\displaystyle w = x + yi$

    So:

    $\displaystyle z = - \frac{\sqrt{3}+3}{2} + \frac{\sqrt{3}-1}{2}i$

    $\displaystyle w = \frac{\sqrt{3}+3}{2} + \frac{\sqrt{3}+1}{2}i$

    Don't know how to prove if i'm right, maybe you can help me again,
    Thanks.
    I don't know what you've done here... You're told $\displaystyle \displaystyle z + w = 1 - i$ and we've found

    $\displaystyle \displaystyle z + w = \frac{\sqrt{3}}{2}r_1 + \frac{1}{2}r_2 + \left(\frac{1}{2}r_1 - \frac{\sqrt{3}}{2}r_2\right)i$.

    Surely that means

    $\displaystyle \displaystyle \frac{\sqrt{3}}{2}r_1 + \frac{1}{2}r_2 = 1$ and $\displaystyle \displaystyle \frac{1}{2}r_1 - \frac{\sqrt{3}}{2}r_2 = -1$.

    Solve these equations simultaneously for $\displaystyle \displaystyle r_1$ and $\displaystyle \displaystyle r_2$.
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