a-2-i-b-a-1-b-1-i-a-2-b-2-i-Powers-of-complex-numbers-

Question Number 9661 by geovane10math last updated on 23/Dec/16

a) 2^{i} =

b) {(a_{1} + b_{1} i)}^{a_{2} + b_{2} i} =

P o w e r s o f c o m p l e x n u m b e r s ? ? ?

Commented by prakash jain last updated on 23/Dec/16

a^{x} = e^{x \ln a}

2^{i} = e^{i \ln 2} = \cos (\ln 2) + i \sin (\ln 2)

Commented by FilupSmith last updated on 23/Dec/16

{(a + b i)}^{x + y i}

{(a + b i)}^{x + y i} = {(a + b i)}^{x} {(a + b i)}^{y i}

= {(a + b i)}^{x} e^{y i \ln (a + b i)}

= {(a + b i)}^{x} {\cos (y \ln {a + b i}) + i \sin (y \ln {a + b i})}

z = a + b i \Rightarrow z^{x + y i} = z^{x} (\cos (y \ln (z)) + i \sin (y \ln (z)))

Commented by geovane10math last updated on 23/Dec/16

T h a n k s