Ken T.

# Functions of Complex Variable

(a) If cos(Θ+i*Φ)=R(cosα+i*sinα),prove that e=sin(Θ-α)/sin(Θ+α)

(b) if u=log(tan(Π/4+Θ/2)),prove that Θ=-i*log(tan( (Π/4) + (i*u)/2))

Bobosharif S.

Did you prove (b)
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01/18/18

Ken T.

I tried but i could not.
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01/18/18

Bobosharif S.

Alright. It would be good to know where did you get  stuck.
Anyway, you get
eu=tan[π/4+Θ/2]
acrtan(eu)=π/4+Θ/2
acrtan(eu)-arctan(1)=Θ/2 <--- (1)
I replaced π/4 by acrtan(1).

There is such formula.
arctan(x)-arctan(y)=arctan[(x-y)/(1+xy)]. (f1)
By using (f1) from (1) we get
arctan[(eu-1)/(eu+1)]= Θ/2. (2)

Now there is another formula
arctan(z)=(1/2i)*log[(1+iz)/(1-iz)]. (f2)
Now plug in in (2) then simplify it..., inside log() you will get tan().
Hope it helps, try..

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01/18/18

Bobosharif S.

In
arctan[(eu-1)/(eu+1)]= Θ/2. (2)
(eu-1)/(eu+1)=(eu/2-eu/2)/(eu/2+eu/2)=tanh(u/2).
Now we use (f2), that is
arctan(tah(u/2))=(1/2i)*log[(1+i tanh(u/2))/(1-i tanh(u/2)]=
=|itanh(x)=tan(ix), therefore|=
=(1/2i)*log[(1+tan(i u/2)/(1- tan(i u/2))]=
=(1/2i)*log[(tan(π/4)+tan(i u/2)/(tan(π/4)- tan(i u/2))]=

=|remind that tan(a+b)=(tan(a)+tan(b))/(tan(a)- tan(b))|=
=(1/2i)*log[tan(π/4+i u/2)].
Basically it is done. But somewhere I'm missing a minus sign (-).
If you look carefully, I'm sure you can find it.
Good luck. Don't forget to write me, when you done!

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01/18/18

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