Question

help
(2+4i)^i

Answer 1

stuff it [the inner bit] is polar complex and it should all fall out nicely

Answer 2

don't understand can u solve it im a visual learner

Answer 3

@IrishBoy123

Answer 4

Thinking :)) simmer down

Answer 5

Well I don't really see this one working out... "nicely" per say...
Lemme show ya what I'm thinking...

Answer 6

\[\large\rm z=(2+4i)^i\]We can rewrite this in some complex form,\[\large\rm z=\left(r e^{i \theta}\right)^i\]and proceed from there.

Answer 7

Oh you went offline... uh ok..

Answer 8

I would procede from here...\[\Large z=e^{i\log(2+4i)}\]

Answer 9

\[\large n^i =\cos(\ln (n))+i\sin(\ln (n))\]

and so:
\[\large (2+4i)^i = (\sqrt{20} \quad e^{i\theta} )^i = \sqrt{20}^i (e^{- \theta} ) \\ = \large (e^{- \theta} ) \left( \cos(\ln \sqrt{20})+i\sin(\ln \sqrt{20}) \right)\]

and \(\large \theta = arctan (2)\)



but, no, this is hardly "nice". mea culpa.

Answer 10

retriceuming it's even right .... tssshhh

Answer 11

i * i = -1

Answer 12

i is gone

-1 appears

Answer 13

that's one of the answers

Answer 14

how do u get the angle

Answer 15

angle = arctan(b/a) ?

Answer 16

@IrishBoy123

Answer 17

\[\large\rm \theta=\arctan\left(\frac{4}{2}\right)\]ya that's your angle,\[\large\rm \theta=\arctan\left(2\right)\]