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    i dont get questions 6.6 and 6.7, for those that have the booklet...otherwise:

    6.6)find all the complex roots of
    a) cosh z =1
    b)sinh z= 1
    c)e^z = -1
    d)cos z=sqrt 2

    6.7) Show:
    w=z+c/z, maps circle |z|=1 uin the z plane into an ellipse in the w plane and find its eqtn

    spent 15 mins or so on each question and gave up
    grateful for your help,
    cheers:cheers:

    phil
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    6.6)find all the complex roots of
    a) cosh z =1
    cosh(a+ib)=1
    cosha.coshib+sinha.sinhib=1
    cosha.cosb+isinha.sinb=1
    Clearly:
    cosha.cosb=1 and sinha.sinb=0
    If sinha=0 then a=0 so cosha=1 and so cosb=1 giving b=2kpi.
    If sinb=0 then b=kpi and cosb=+-1 but we require cosb to be positive as cosha is positive so b=2kpi and a=0.
    Hence the complex roots are 0+i(2kpi) where k \in Z (0, \pm 1, \pm 2, \pm 3...)

    b)sinh z= 1
    Very similar to the above but using the correct identity for sinh(A+B).

    c)e^z = -1
    cosz+isinz=-1
    Clearly cosz=-1 and sinz=0.
    Hence z=(2k+1)pi where k \in Z (0, \pm 1, \pm 2, \pm 3...)


    d)cos z=sqrt 2
    Use z=a+ib, expand and equate real and imaginary coefficients as in (a) or (b). Obviously remember to use the correct identities as you're dealing with a trigonometric rather than hyperbolic function.

    6.7) Show:
    w=z+c/z, maps circle |z|=1 uin the z plane into an ellipse in the w plane and find its eqtn
    Clearly attempts to rearrange the formula to find z in terms of w are going to give us a quadratic and thus things will get a little messy.
    Instead, let z=x+iy then x^2+y^2=1.
    w=x+iy+c/(x+iy)
    w=x+iy+c(x-iy)/(x^2+y^2)
    w=x+iy+c(x-iy)/1
    w=x+iy+cx-icy
    w=(x+cx)+i(y-cy)
    w=x(c+1)+iy(1-c).
    If w=u+iv then u=x(c+1), v=y(1-c).
    Clearly x=u/(c+1) and y=v/(1-c).
    You know x^2+y^2=1 so u^2/(c+1)^2+v^2/(1-c)^2=1.
    c is a constant so u^2/(1+c)^2+v^2/(1-c)^2=1 is of the form u^2/a^2+v^2/b^2=1 where a and b are constants so the locus is indeed an ellipse.
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    thanks for that!
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    (Original post by Phil23)
    thanks for that!
    No problem but I made a minor correction. Sorry about the confusion.
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    i dont get the english of this problem:

    A circular cone has its vertex at the origin and its axis in the direction of the unit vector a. The half angle at the vertex is alpha.

    what is a hlaf angle? - is it the angle that encompasses the two sides of the cone or is it half of this...i.e. the angle between the axis of symmetry and one side?
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    (Original post by Phil23)
    i dont get the english of this problem:

    A circular cone has its vertex at the origin and its axis in the direction of the unit vector a. The half angle at the vertex is alpha.

    what is a hlaf angle? - is it the angle that encompasses the two sides of the cone or is it half of this...i.e. the angle between the axis of symmetry and one side?
    It's that
 
 
 
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