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    (Original post by Zacken)
    So what's the problem? The argument of a complex number raised to a power is the power multiplied by the argument.
    Oh is that it then ? I think because it was in the form arg(...) i was getting confused, thanks for clearing it up.
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    (Original post by coolguy123456)
    Oh is that it then ? I think because it was in the form arg(...) i was getting confused, thanks for clearing it up.
    Yes, that's it. arg(...) just means "what is the argument of ...".
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    (Original post by Zacken)
    Yes, that's it. arg(...) just means "what is the argument of ...".
    Yeah i forgot that the argument is just the angle and in the whole z^n = r(cos ... etc thing where you multiply angle (realise it is the argument) with the power. Thanks
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    (Original post by coolguy123456)
    yeah i get that but you went from arg((0.5W)^1/3) to arg (0.5W/3) ??
    no I went from arg((0.5w)^(1/3)) to 1/3*arg(0.5w)
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    (Original post by jazztheman)
    How did u go from second line to third line
    rearrange w to make z the subject, then subsitute this into the arg(z)
    I dont know what the problem with this question is to be honest, you should all know logs by now
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    (Original post by coolguy123456)
    Yeah i forgot that the argument is just the angle and in the whole z^n = r(cos ... etc thing where you multiply angle (realise it's not the argument) with the power. Thanks
    No worries.
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    (Original post by kennz)
    rearrange w to make z the subject, then subsitute this into the arg(z)
    I dont know what the problem with this question is to be honest, you should all know logs by now
    Yeh makes sense sorry haha we all have those dumb moments thanks again good luck for the exam
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    (Original post by kennz)
    arguments work like logs
    (Original post by kennz)
    arguments work like logs
    (Original post by kennz)
    arguments work like logs
    Wow so they do 😯 😯 feel slightly silly for not realising that earlier, it's a really good way of thinking about it.

    Anyways... good luck for Wednesday everyone. I'll be on here straight afterwards haha x
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    For anybody interested, I'll be doing model solutions the day after.
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    (Original post by Zacken)
    For anybody interested, I'll be doing model solutions the day after.
    Cool
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    (Original post by Zacken)
    For anybody interested, I'll be doing model solutions the day after.
    u beta get everything rite. dnt get stuk on complex numbaz
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    How would one simplify \dfrac{isinx}{2-2cosx}
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    (Original post by Zacken)
    For anybody interested, I'll be doing model solutions the day after.
    Will it be paper included or same style like you did with other modules (ie C2)
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    (Original post by edothero)
    Will it be paper included or same style like you did with other modules (ie C2)
    Same like I did with C2. Wish I could include the paper but copyright. :/
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    (Original post by Zacken)
    Same like I did with C2. Wish I could include the paper but copyright. :/
    No worries, I hope I can remember my answers :laugh:
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    (Original post by edothero)
    How would one simplify \dfrac{isinx}{2-2cosx}
    Half angle formulae
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    (Original post by Student403)
    Half angle formulae
    Or double angle formulae where the double angle is theta
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    (Original post by Student403)
    Or double angle formulae where the double angle is theta
    ya took a while sinx = 1/2sin(x/2)cos(x/2) and cosx= 1-2sin(x/2)
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    Quite a good question actually, if anyone wants to try.

    w = \dfrac{1}{z-1}

    given that z=e^{i\theta}
    Show that

     w = -\frac{1}{2}-\frac{1}{2}icot(\frac{1}{2} \theta)
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    (Original post by edothero)
    Quite a good question actually, if anyone wants to try.

    w = \dfrac{1}{z-1}

    given that z=e^{i\theta}
    Show that

     w = -\frac{1}{2}-\frac{1}{2}icot(\frac{1}{2} \theta)
    Yeah it was cool
 
 
 
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