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(Thinking)(Thinking)(Thinking)

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- Jul 11th 2013, 06:31 AMspacemenoncomplex in exponetial form
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(Thinking)(Thinking)(Thinking)

- Jul 11th 2013, 07:00 AMPlatoRe: complex in exponetial form
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? - Jul 11th 2013, 07:13 AMspacemenonRe: complex in exponetial form
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)

- Jul 11th 2013, 07:34 AMPlatoRe: complex in exponetial form
- Jul 11th 2013, 05:24 PMspacemenonRe: complex in exponetial form
sqrtz =z1 and z2

and arg z1 =pi/3 which is a positive angle so nxt root argument shld be pi/3 -pi =-2pi/3 ,still i dont get hw u got 4pi/3

pls check my wrking pdf attached - Jul 12th 2013, 07:30 AMHallsofIvyRe: complex in exponetial form
- Jul 14th 2013, 02:59 AMspacemenonRe: complex in exponetial form
sure, can will you point out were is the mistake,in the pdf which i have uploaded ,can you point me to any error in my working

- Jul 14th 2013, 04:35 AMPlatoRe: complex in exponetial form
@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. - Jul 15th 2013, 08:22 AMspacemenonRe: complex in exponetial form
ok thank you for the help

- Aug 4th 2013, 08:37 AMspacemenonRe: complex in exponetial form
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 - Aug 4th 2013, 08:58 AMPlatoRe: complex in exponetial form
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. - Aug 4th 2013, 04:54 PMspacemenonRe: complex in exponetial form
so instead of exp(i 4(pi)/3) i can write the answer as exp(i(-2(pi)/3),that is instead of exp(i 240) i give the answer has exp(i (-120)) ,

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