So I recently found out that
sin(a+bi)=isinh(b−ai)
cos(a+bi)=cosh(b−ai)
and that
sinh(x)=2ex−e−x
cosh(x)=2ex+e−x
Since
ba+ab=aba2+b2
ba−ab=aba2−b2
and
(ba)−x=(ab)x
Re-arranging the equations for sinh(x) and cosh(x) gives us
sinh(x)=2exe2x−1
cosh(x)=2exe2x+1
Substituting these into sin(a+bi) and cos(a+bi) when x=b−ai gives us
sin(a+bi)=i2eb−aie2(b−ai)−1
cos(a+bi)=2eb−aie2(b−ai)+1
I'm going to try and expand and substitute by using
ea+bi=ea(cos(b)+isin(b))
Let cosθ+isinθ= cis θ. So
ea+bi=ea⋅cis(b)
Next we substitute this new equation in to get
sin(a+bi)=i2eb⋅cis(−a)e2b⋅cis(−2a)−1
cos(a+bi)=2eb⋅cis(−a)e2b⋅cis(−2a)+1
So what does cis(a+bi) equal
Substituting in the two equations gives us
cis(a+bi)=2eb⋅cis(−a)e2b⋅cis(−2a)+1+i(i2eb⋅cis(−a)e2b⋅cis(−2a)−1)
That looks unfriendly, let's simplify it
cis(a+bi)=eb⋅cis(−a)1
(Go through the steps, it works)
This can be simplified even further
cis(a+bi)=ebcis(a)
P.s. (cosθ+isinθ)n=cosθn+isinθn
Easy Math Editor
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Top NewestVery nice note. Learned a lot!
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cis(a+bi)=ei(a+bi)=eia−b=ebeia=ebcis(a)
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I did consider that but I wanted to make it as complex as possible so it would be fun to work out.
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Lol..Nice note!!
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Although trigonometric functions can operate complex numbers, I think there is no complex angle but only real angles. Just like a doctor who can save lives, not all lives being saved are human being. Do you agree with me?
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