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Find a solution for the ff: (without the figure i) (in terms of a,b,c and d)
(a + bi)^(c + di)

This might help:
I have discovered that
i^i = e^(-π/2)

E.g.
(2 + 3i)^(1 - 2i)

^_^

2006-09-21 01:28:34 · 3 answers · asked by kevin! 5 in Science & Mathematics Mathematics

3 answers

This deals with complex numbers and not imaginary
let a= r cos t
b = r sin t

r = sqrt(a^2+b^2)
t= tan ^-1 (b/a)
now (a+ib) = re^(it)
now (a+ib)^(c+id) = (re^(it))^(c+id)
= r^(c+id)*e^(cit-dt)
= r^c e^(cit-dt) r^id

for r = e^(ln r)
r^(id) = (e^(in r)^id)
so result = r^ce^(-dt)e^(cit)e^(idln(r))
= r^ce^(-dt)e^(i(ct-d ln r)
=r^ce^(-dt)(cos(ct-d ln r) + i sin(ct-d ln r))
where r = sqrt(a^2+b^2)
and t = tan^-1(b/a)

2006-09-21 02:23:13 · answer #1 · answered by Mein Hoon Na 7 · 0 0

the answer is 9.896ac π
~an imaginary answer to an imaginary Question~
lol
^_-

2006-09-21 08:33:01 · answer #2 · answered by Pauli :) 6 · 0 0

6.5

2006-09-21 08:30:08 · answer #3 · answered by jemz 1 · 0 0

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