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Hence express tan (pi/12) in surd form.

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Let z= √3 -i

    and w= 1+i

    i)calculate zw in "a+ib"form

    ii)Write z and w in polar form and thus write zw in polar form

    iii)Hence express tan (pi/12) in surd form.



    2. Relevant equations



    3. The attempt at a solution

    (√3 -i)(1+i)

    = (√3+1)+(√3i-i)

    ii)

    |z|=2 → Arg(z)=tan^-1(√3/-1) = -pi/3

    so z = 2*e^-i(pi/3)

    |w|= √2 → Arg(w)=pi/4

    so w=√2 *e^i(pi/4)

    so zw= 2√2*e^i(pi/4 -pi/3)

    = √8*e^-i(pi/12)


    iii) this is the part I'm having trouble with...

    so I think I should convert it into "a +ib" form

    so |zw| = √8 → Arg(zw)=-pi/12

    but then Im stuck because there is no exact result for -pi/12

    ????
     
  2. jcsd
  3. Apr 13, 2012 #2

    tiny-tim

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    hi charmedbeauty! :smile:

    (have a pi: π and try using the X2 button just above the Reply box :wink:)

    hint: if eiπ/12 = p + iq, what is tan(π/12) ? :wink:
     
  4. Apr 13, 2012 #3

    Curious3141

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    Check the math here very carefully, then redo what you did, then apply TinyTim's hint.
     
  5. Apr 13, 2012 #4
    Im pretty sure the math here checks out?

    unless it should be 2e^i(-π/3)

    but im still stuck on tinytims hint

    I know that tan =sin/cos

    and 2√2 e^i(π/12)

    = 2√2[cos(π/12)+isin(π/12)]

    but im stumped to see how you can relate these?

    and as I understand it I shouldn't be trying to find the real and imaginary part of Arg(z)=π/12 and mod of √8 because that is not going to turn out nice.

    unless I divided the terms by cos(π/12) to get a term with tan but I still think thats wrong?
     
  6. Apr 13, 2012 #5
    [itex]\theta=\arctan \left( {\frac {q}{p}} \right)[/itex], so [itex]\tan \left( \theta \right) ={\frac {q}{p}}[/itex]

    theta = Pi/12

    You already have q and p.

    The other (longer) method is to use the identity (sin x)/(cos x) = tan x

    Since Pi/12 = 15 degrees. (sin 45 - 30) = sin(45)cos(30)...etc.etc. substitute in the exact values in surd form, and then divide. The answer works out to the same.
     
  7. Apr 13, 2012 #6

    Curious3141

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    You're supposed to take the arctangent of the Imaginary part/Real part.

    The imaginary part here is -1, the real part is √3.

    So the argument is arctan (-1/√3) = ??? :smile:

    Right, once you get to this stage, you compare it to the a + bi form you worked out earlier.

    So here you're comparing 2√2cos(π/12) + i*2√2sin(π/12) to (√3+1)+i*(√3-1)

    Equate the real and imaginary parts.

    You can then get a surd expression for the cosine, and another for the sine. Just divide the sine by the cosine (as you know), and simplify (do the usual rationalisation of the denominator), and you're done! :biggrin:
     
  8. Apr 13, 2012 #7
    ok thanks I see what you mean about check the math... I should have picked up something was wrong when I had the -π/12 instead of π/12

    ok,, I should be right from here thanks for tips, your help really helps!!.
     
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