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# Problems with complex functions/equations watch

1. 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?!

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?

Robbie
2. (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?!
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:

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
Hence and so
(Remember that sine and cosine are periodic with period 2pi)
3. 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.
4. 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 Thus iff for some integer k.
It might help you to to see this if you write out both sides of your equation in the form , find an expression for z in polar form and hence find z in the form a+ib.

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Updated: November 21, 2005
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