Question

http://www.wolframalpha.com/input/?i=%28sqrt3%2F2+-+1%2F2+i%29^2
can any one explain why?....or give the extended solution?

Answer 1

\[(\sqrt{ \small {3\over 2}} - \small{{1\over 2}} i) (\sqrt{ \small {3\over 2}} - \small{{1\over 2}} i)\]

Answer 2

USe Euler's formular, \[re ^{i \theta}=r[\cos \theta+isin \theta]\]

Answer 3

earth citizen...what the hell r u talking about???????????????

Answer 4

\[z ^{n}=(re ^{i \theta })^{n}=r ^{n}[\cos(n \theta)+isin(n \theta)]\]

Answer 5

ok...can u demonstrate that with the numbers in my problem? cause i think you've gone way too far from basic algebra

Answer 6

(sqrt3/2 - 1/2 i)^2=(sqrt3/2 - 1/2 i)*(sqrt3/2 - 1/2 i)

Answer 7

let, \[z=(\sqrt{3}-i)/2\]

Answer 8

(sqrt3/2)*(sqrt3/2)-(sqrt3/2)*(1/2 i)-(1/2 i)*(sqrt3/2)+(1/2 i)(1/2 i)

3/4-(sqrt3/2 i)-1/4

Answer 9

i'm not giving you a medal for that...it's just more confusing

Answer 10

kk

Answer 11

\[((\sqrt{3}-i)/2)^{2}= (1-i \sqrt{3})/2\]

Answer 12

dude...have u been drinking?

Answer 13

no, i don't drink

Answer 14

\[r <\theta\]\[r=1, \theta=\arg(z)=\tan^{-1} (-\sqrt{3})=-60^{o}\]

Answer 15

i'm sorry Spyros, this is not basic algebra per se it's complex analysis

Answer 16

when you multiply out the square you'll get (1-isqrt(3))/2 then convert that to polar form