Question

Write e^4-2i in the a+bi form. Thank you

Answer 1

\[e ^{(4-2i)}\] in the a+bi form.

Answer 2

HINT: \(\Large e^{xi} = \cos(x) + i\sin(x)\)

Answer 3

Better Hint: \(\Large e^{a+bi} = e^ae^{bi} = e^a(\cos(b) + i\sin(b))\)

Answer 4

@Haleo

Answer 5

Can you handle this problem now with my hints?

Answer 6

@geerky42 , I will now try. I need a minute.

Answer 7

ok

Answer 8

@geerky42, from my stuffing terms in the formula you gave (how is it called, I didn't go through it, I guess), I get the following:

\[e ^{(4-2i)}= e ^{4}* e ^{-2i} = e ^{4}(\cos(-2) + i \sin (-2))\]

But I don't feel it's right.

Answer 9

Please, I need to finish calculating this! Help me, please.

Answer 10

to put it into a+bi form, you should distribute the e^4
you can simplify cos(-2) to cos(2)
and simplify sin(-2) to -sin(2)

Answer 11

According to @ParthColi,
a= \[e ^{4}\cos(-2) = e ^{4}\cos(2)\]

and b = \[e ^{4}\sin(-2) = -e ^{4}\sin(2)\]

But I don't understand how/why do the answers differ (both are -2) and how do get from that to the a+bi form.

Answer 12

@phi

Answer 13

you wrote
**\[ e ^{(4-2i)}= e ^{4}* e ^{-2i} = e ^{4}(\cos(-2) + i \sin (-2)) \]
which is the same as
\[ e ^{4} \cos(-2) + i e ^{4}\sin (-2) \]

it is well-known (to those who know trig well) that cos(-2)= cos(2)
and sin(-2) is the same as -sin(2)

so it looks a little nicer to write it as
\[ e ^{4} \cos(2) - i\ e ^{4}\sin (2) \]
notice the i in the 2nd term: you have the form a+b i

you could write this as a decimal
\[ \approx -22.72 -49.65 i \]
but that is only an approximation

Answer 14

pellet, it was correct when I tried to fondle a bit with the digits...the numbers seem so irrational that I thought it was a blah....Thank you for you help.

Answer 15

hahah...they changed pellet to pellet...

Answer 16

Oh, lol!