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Mathematics: Post your doubts here!

A

_Ahmad

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sara kamal said:
http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_qp_31.pdf
please anyone?
explain q5 part 2
Click to expand...

z=9 e^i(1/3)

z=9(e^0(cos(1/3)+i sin(1/3))

square root of z=(9 (e^0(cos(1/3)+i sin(1/3)) )^1/2

=9^(1/2)( cos(0.5*(1/3 π))+i sin(0.5*(1/3 π)) )

=+-3(cos(1/6 π) + i sin(1/6 π))

one square root is =3(cos(1/6 π) + i sin(1/6 π))
= (3√3/2)+i (3/2) (open the bracket)
r=3
θ=1/6 π
so 3 e^i (1/6) π

other root = - 3(cos(1/6 π) + i sin(1/6 π))
= - (3√3/2)- i (3/2)
r=3
θ = (1/6 π)
θ= (1/6 π)-π = -(5/6)π
so 3 e^i (-5/6) π
 
A

_Ahmad

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The Sarcastic Retard

The Sarcastic Retard

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Ram97 said:
http://studentbounty.com/pastpapers/Cambridge International Examinations (CIE)/International AS and A Level/Mathematics (9709)/2010 Jun/9709_s10_qp_12.pdf
Q.11 part iii and v
Click to expand...
Thanks to abcde and rookz didlu :p
iii)f(x) = k
=> 4 - 3 sin x = k
=> sin x = (4 - k)/3
sin x has values ranging from 1 to -1.
=> -1 ≤ (4 - k)/3 ≤ 1
=> k ≤ 7, k ≥ 1.
So for the function to have no solution, the ranges of k are: k > 7 and k < 1.

v)
g inverse we get is sin inverse (4-x/3) so g inverse 3 = sin invers 1/3 and that value comes out to be o.34 and thats outta given domain.
x = pi - 0.340 = 2.80 u do this coz limit is b/w 0.5 pi to 1.5 pi hence its found in 2nd quadrant
 
tiki-taka

tiki-taka

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Jennifer4678 said:
Can someone please help me with 5(ii) :confused:
Click to expand...
done by _Ahmad

z=9 e^i(1/3)

z=9(e^0(cos(1/3)+i sin(1/3))

square root of z=(9 (e^0(cos(1/3)+i sin(1/3)) )^1/2

=9^(1/2)( cos(0.5*(1/3 π))+i sin(0.5*(1/3 π)) )

=+-3(cos(1/6 π) + i sin(1/6 π))

one square root is =3(cos(1/6 π) + i sin(1/6 π))
= (3√3/2)+i (3/2) (open the bracket)
r=3
θ=1/6 π
so 3 e^i (1/6) π

other root = - 3(cos(1/6 π) + i sin(1/6 π))
= - (3√3/2)- i (3/2)
r=3
θ = (1/6 π)
θ= (1/6 π)-π = -(5/6)π
so 3 e^i (-5/6) π
 
J

Jennifer4678

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tiki-taka said:
done by _Ahmad

z=9 e^i(1/3)

z=9(e^0(cos(1/3)+i sin(1/3))

square root of z=(9 (e^0(cos(1/3)+i sin(1/3)) )^1/2

=9^(1/2)( cos(0.5*(1/3 π))+i sin(0.5*(1/3 π)) )

=+-3(cos(1/6 π) + i sin(1/6 π))

one square root is =3(cos(1/6 π) + i sin(1/6 π))
= (3√3/2)+i (3/2) (open the bracket)
r=3
θ=1/6 π
so 3 e^i (1/6) π

other root = - 3(cos(1/6 π) + i sin(1/6 π))
= - (3√3/2)- i (3/2)
r=3
θ = (1/6 π)
θ= (1/6 π)-π = -(5/6)π
so 3 e^i (-5/6) π[
Click to expand...


Thanks!! But I still don't quite understand the front portion.. How did you get that second equation with e^0 ? o_O
 
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