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    Im asked in a question to show that for all complex numbers z/w to show that:

    cos(z+w) = coszcosw -sinzsinw...

    Its an identity... Im not sure how I can SHOW that?! :confused:

    Additionally it asks me to deduce that cos^2z+sin^2z = 1, once again its an identity... how can it be DEDUCED??

    How do you find all solutions to
    sin(z)=0??

    My effort involves the following...
    sin^-1 of both sides.. would give z = 0, pi, 2pi... clearly not right... Help

    Also I need to solve the following equation:
    e^z = e^(2+i)... ln both sides, gives
    z = 2+i... It can't be that easy, what am I missing?

    Please help,
    Robbie
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    (Original post by RobbieC)
    Im asked in a question to show that for all complex numbers z/w to show that:

    cos(z+w) = coszcosw -sinzsinw...

    Its an identity... Im not sure how I can SHOW that?! :confused:
    Being an identity just means its true for all values, not that it can't be deduced from more already known facts.
    Proceed by noting that:
    cos(z+w)=Re (e^{i(z+w)})\\

=Re (e^{iz}e^{iw})\\

=Re ((\cos (z)+i\sin (z))(\cos (w)+i\sin (w)))

    Additionally it asks me to deduce that cos^2z+sin^2z = 1, once again its an identity... how can it be DEDUCED??
    Use a similar idea to above,

    How do you find all solutions to
    sin(z)=0??

    My effort involves the following...
    sin^-1 of both sides.. would give z = 0, pi, 2pi... clearly not right... Help
    Let z=a+ib, expand and use the relations with hyperbolics as necessary to find all solutions.

    Also I need to solve the following equation:
    e^z = e^(2+i)... ln both sides, gives
    z = 2+i... It can't be that easy, what am I missing?
    Let z=a+ib.
    Then e^a e^{ib}=e^{2}e^{i}
    Hence a=2 and e^{ib}=e^{i} so b=1+2k\pi
    (Remember that sine and cosine are periodic with period 2pi)
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    How do you find all solutions to
    sin(z)=0??

    My effort involves the following...
    sin^-1 of both sides.. would give z = 0, pi, 2pi... clearly not right... Help


    Let z=a+ib, expand and use the relations with hyperbolics as necessary to find all solutions.

    Expand what? Surely that would just mean a + ib = 0,pi,2pi... That doesnt add anything to what I already had except some random unknowns... If Im completely off track could you be more explicit as to what im meant to be doing?

    Quote:
    Also I need to solve the following equation:
    e^z = e^(2+i)... ln both sides, gives
    z = 2+i... It can't be that easy, what am I missing?


    Let z=a+ib.
    Then
    Hence and so
    (Remember that sine and cosine are periodic with period 2pi)
    Im not sure what cosine/sine have to do with this question, there are only exponentials, arent there... I dont follow beyond the first line, and what you were doing doesn't seem to lead toward a solution.
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    Expand what? Surely that would just mean a + ib = 0,pi,2pi... That doesnt add anything to what I already had except some random unknowns... If Im completely off track could you be more explicit as to what im meant to be doing?
    We have:
    sin(a+ib)=0
    sinacosib-cosasinib=0
    (sina)(coshb)-i(cosa)(sinhb)=0
    Equating real and imaginary parts:
    (1) (sina)(coshb)=0
    (2) (cosa)(sinhb)=0
    From (1), since (coshb)>0 we see sina=0 so a=kpi for some integer k.
    From (2), since coskpi is never zero we see that sinhb=0. Hence you have shown that b=0.
    Hence z=kpi are the only solutions.

    --------------

    Im not sure what cosine/sine have to do with this question, there are only exponentials, arent there... I dont follow beyond the first line, and what you were doing doesn't seem to lead toward a solution.
    Recall that e^{i\theta}=cos\theta + i\sin \theta Thus e^{i\theta}=e^{i\phi} iff \theta=2k\pi + \phi for some integer k.
    It might help you to to see this if you write out both sides of your equation in the form r(cos\theta + i\sin \theta ), find an expression for z in polar form and hence find z in the form a+ib.
 
 
 

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