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    z = 6 ( Cos (5π/6) + isin(5π/6) )
    w = 4 (Cos (-π/4) +isin(-π/4) )

    Write the following complex numbers in the form r (cosθ+isinθ),
    where r > 0 and -π < θ ≤ π

    i) z/w
    ii) w/z
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    (Original post by 98701)
    z = 6 ( Cos (5π/6) + isin(5π/6) )
    w = 4 (Cos (-π/4) +isin(-π/4) )

    Write the following complex numbers in the form r (cosθ+isinθ),
    where r > 0 and -π < θ ≤ π

    i) z/w
    ii) w/z
    What have you tried? Post your working.
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    (Original post by 98701)
    z = 6 ( Cos (5π/6) + isin(5π/6) )
    w = 4 (Cos (-π/4) +isin(-π/4) )

    Write the following complex numbers in the form r (cosθ+isinθ),
    where r > 0 and -π < θ ≤ π

    i) z/w
    ii) w/z
    i . your modulus is just 6 / 4 which is r in modulus arg form
    your argument is 5n/6 - n/4

    ii. same method just switch numbers around
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    (Original post by Tbarker1)
    i . your modulus is just 6 / 4 which is r in modulus arg form
    your argument is 5n/6 - n/4

    ii. same method just switch numbers around
    That is what i got but is 5n/6 - n/4 a principle argument?
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    (Original post by 98701)
    That is what i got but is 5n/6 - n/4 a principle argument?
    yeah its the argument you place after cos( and isin(
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    (Original post by Tbarker1)
    yeah its the argument you place after cos( and isin(
    Ok, Thanx
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    Another way is to convert z and w into exponential form and (z/w) and (w/z) can then be easily tranformed back into the modulus and argument form.

    Or note that:

     \displaystyle \frac{w}{z} = \frac{ \lvert w \rvert }{ \lvert z \rvert} \big( \cos \left[ \mathrm{arg}(w) - \mathrm{arg}(z) \right] + i \sin \left[ \mathrm{arg}(w) - \mathrm{arg}(z) \right] \big) .

    Can you take this forward?
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    (Original post by simon0)
    Another way is to convert z and w into exponential form and (z/w) and (w/z) can then be easily tranformed back into the modulus and argument form.

    Or note that:

     \displaystyle \frac{w}{z} = \frac{ \lvert w \rvert }{ \lvert z \rvert} \bigg( \cos \big( \mathrm{arg}(w) - \mathrm{arg}(z) \big) + i \sin \big( \mathrm{arg}(w) - \mathrm{arg}(z) \big) \bigg) . .

    Can you take this forward?
    should be able to
    Thank you \(^o^)/
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    i) 6/4(Cos(13/12 π)+iSin(13/12 π))
    θ > π so -2π to give:
    6/4(Cos(-11/12 π)+iSin(-11/12π))

    ii)4/6(Cos(-13/12 π)+iSin(-13/12 π))
    θ < π so + 2π to give:
    4/6(Cos(11/12 π)+iSin(11/12 π))
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    (Original post by physconomics)
    i) 6/4(Cos(13/12 π)+iSin(13/12 π))
    θ > π so -2π to give:
    6/4(Cos(-11/12 π)+iSin(-11/12π))

    ii)4/6(Cos(-13/12 π)+iSin(-13/12 π))
    θ < π so + 2π to give:
    4/6(Cos(11/12 π)+iSin(11/12 π))
    Thank you :^_^:
 
 
 
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