Question

Write the following numbers in the form re^(iθ) and in the form a + ib:
(1+i)^20

Answer 1

let x=1+i
x^2=1+2i+i^2=2i
and
\[x^{20}=x^{2 \cdot 10} =(x^2)^{10}=(2i)^{10}\]

Answer 2

can you simplify it from there ?

Answer 3

Yeah, the part I don't understand us writing it in re^(iθ) form

Answer 4

re^(i*theta)

Answer 5

so you got what after computing (2i)^10?

Answer 6

because we can use that to write in re^{i theta) form

Answer 7

-1024

Answer 8

cos equals -1 for what angle?

Answer 9

cos(pi)=-1

Answer 10

-1024(1+0i)=1024(-1+0i)

Answer 11

yes

Answer 12

so \[-1024(1+0i)=1024(-1+0i)=1024(\cos(\pi)+i \sin(\pi))\]

Answer 13

and guess what that equals :)

Answer 14

-1024

Answer 15

well in re^(i theta) form

Answer 16

remember in earlier threads we said
\[e^{i \theta}=\cos(\theta)+i \sin(\theta)\]

Answer 17

1024e^(pi*i)

Answer 18

bring

Answer 19

bingo

Answer 20

I think I'm getting tired
i keep typing wrong things

Answer 21

bring lol

Answer 22

Thank you for all your help tonight!
Its okay lol me too.

Answer 23

did this all make sense

Answer 24

i sorta used the method we used on last thread to find that binomial to the 20th power

Answer 25

i like it :)

Answer 26

when it can be used anyways

Answer 27

Yeah it did make sense! it will help a lot for the other problems like it :)

Answer 28

so the longer way
\[(1+i)^{20} =(\sqrt{2}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i))^{20}=(\sqrt{2})^{20}(\cos(\frac{\pi}{4})+\sin(\frac{\pi}{4})i)^{20} \\ = 2^{10}(\cos(20 \frac{\pi}{4})+i \sin(20 \frac{\pi}{4})) =2^{10}(\cos(5\pi)+i \sin(5 \pi))=2^{10}(-1+0)\]

Answer 29

ick

Answer 30

i thought the first way was rather snappy

Answer 31

yes that one was more the bread pudding way
and this other way was like icky odd yogurt

Answer 32

then again i guess
\[\huge 1+i=\sqrt{2}e^{\frac{\pi}{4}i}\] will work nicely too

Answer 33

no not really
since \(\frac{\pi}{4}\times 20=5\pi\) it is not so bad

Answer 34

i have to admit i would never have thought to compute
\((1+i)^2\) first