Let's "prove" that every complex number z
If z=0 it's obvious. For all other complex numbers z=r*e^(θi), where r is a real number, and i=√-1.
Now, z= r*e^(θi)= r*(e^(2πi))^(θ/2π). Now as we know that e^(2πi)=1 we can write z =r*(1)^(θ/2π) → z=r.
What's wrong with this?
exp^(ab)=(exp^a)6b does not apply to complex no.it is only for real numbers.if it apply to complex no then say b is complex
(exp^(a))^b=(cos a +sin a)^b=cos ab + sin ab
quantity ab is complex, so cosine or sine of it is meaningless
exp^(θ*i)=cos θ + i sin θ is not equal to
cos (θ*i) + sin (θ*i) as your reasoning apply