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Further Maths Complex Numbers

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
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.
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
Reply 3
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?
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(
Reply 5
Original post by Tbarker1
yeah its the argument you place after cos( and isin(


Ok, Thanx :smile:
Reply 6
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:

wz=wz(cos[arg(w)arg(z)]+isin[arg(w)arg(z)]). \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?
(edited 6 years ago)
Reply 7
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:

wz=wz(cos(arg(w)arg(z))+isin(arg(w)arg(z))). \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^)/
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 + to give:
4/6(Cos(11/12 π)+iSin(11/12 π))
Reply 9
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 + to give:
4/6(Cos(11/12 π)+iSin(11/12 π))


Thank you :^_^:

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