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    I've been told |Z|= 4 and that arg z = 1/3(pi)?
    I'm stumped, I have to find a and b in a+bi.

    Can anybody please help me?
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    (Original post by robbo9)
    I've been told |Z|= 4 and that arg z = 1/3(pi)?
    I'm stumped, I have to find a and b in a+bi.

    Can anybody please help me?
    If you were given some general complex number a+bi and asked to find the modulus and the argument what would you do? The only difference is now you have letter, not numbers. You should be able to come up with 2 equations you can solve simultaneously.
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    Have you drawn a picture? You can solve it that way.
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    Or you could use the Rec button on your calculator .... you might not recognise the surd that way though.
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    (Original post by robbo9)
    I've been told |Z|= 4 and that arg z = 1/3(pi)?
    I'm stumped, I have to find a and b in a+bi.

    Can anybody please help me?
    a^2+b^2=4^2
    tan\frac{\pi}{3}=\frac{b}{a}
    Solve somultaneously
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    You should know that you can write any complex number in the form r(\cos \theta + i\sin \theta)... and you've been given what r, \theta are (just with different names). So if z=r(\cos \theta + i\sin \theta) = a+ib then what are a and b?

    (Original post by ztibor)
    a^2+b^2=4^2
    tan\frac{\pi}{3}=\frac{b}{a}
    Solve somultaneously
    That's not a good route to take, since it's not sensitive to the sign of the pair (a,b). That is, (-a)^2+(-b)^2=a^2+b^2 and \dfrac{b}{a} = \dfrac{-b}{-a}, and so you'll get two possible solutions, and you can't infer from the equations what the correct sign is.
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    Thanks guys. I got it as A=2 and B=2root3.

    I'm not feeling good about this exam at all, but at least I've got this month.
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    (Original post by nuodai)
    You should know that you can write any complex number in the form r(\cos \theta + i\sin \theta)... and you've been given what r, \theta are (just with different names). So if z=r(\cos \theta + i\sin \theta) = a+ib then what are a and b?


    That's not a good route to take, since it's not sensitive to the sign of the pair (a,b). That is, (-a)^2+(-b)^2=a^2+b^2 and \dfrac{b}{a} = \dfrac{-b}{-a}, and so you'll get two possible solutions, and you can't infer from the equations what the correct sign is.
    Yes, but you have to know that for minus sign the argument would be
    4*pi/3. For pi/3 a>0 and b>0
    HAve you any other problem?
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    Thank you. I have no other problems for now. I can see my self being back though.
 
 
 
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