Pls share ur ideas i am trying to find the square root of e^2/3pi and answer is given in text as e^(pi/3)i,e^(4pi/3),i dont understand hw is possible to get e^4pi/3,i am getting e^2pi/3 pls explain
As with your other post, this one equally hard to read. Please try to learn how to post correctly using grouping symbols to make the exponents clear.
Here is a couple of simple observations.
If $\displaystyle \alpha$ is one square root of a complex number then $\displaystyle -\alpha$ is the other.
For any real number $\displaystyle r$, $\displaystyle (\exp(i\theta))^r=\exp(r\,i\theta))$
Thus it ought be clear to you that $\displaystyle \left[ {\exp \left( {\frac{{i\pi }}{3}} \right)} \right] = {\left[ {\exp \left( {\frac{{2i\pi }}{3}} \right)} \right]^{\frac{1}{2}}}$.
Thus you have one root: $\displaystyle \exp \left( {\frac{{i\pi }}{3}} \right)}$.
What is the other one?
sure i will try my best next time to make it more reader friendly.so u r implying my answer exp(i pi/3) and exp(-2 pi/3) is the sqrt of exp(2 pi/3) then frm were do we get exp(4 pi/3),that one of the sqrt given along with exp(i pi/3)
@spacemenon, What gives you the right to expect us to open a pdf attachment?
Frankly, I will do it. Why don't you learn to post in LaTeX?
You have already been told that if $\displaystyle u~\&~v$ are two principal square roots of $\displaystyle z$ then $\displaystyle \arg(u)=\arg(w)\pm\pi~.$
So, frankly I don't see what you are calling is a mistake. We are not here to teach you the subject matter.
Thus if you need more help, learn to post readable symbols.
why are you adding (pi)/3 to pi to find your next argument iwould have subtracted pi
pi radians - 180 degrees
i got one argument as 60 degrees so the next square root argument should be 60-180= -120 degrees since agrument should be frm -180 to +180, but you have added 180 degrees which will be 240 degrees which outside our range -180 to +180
I request your comment
Whereas $\displaystyle -\pi<\theta\le\pi $ is the usual value for the argument function, there is no hard and fast rule to that effect.
After all, $\displaystyle {240^ \circ } \equiv - {120^ \circ }$.
I will say in programming a computer algebra system to display the complex roots it is a lot easier to do it the way I posted. In fact that is exactly the reason I use that standard.