Can anyone show me how to determine the Arg(2+i) ?

I know that $\displaystyle \theta=sin^{-1}(\frac{1}{\sqrt{5}})$

But I don't know how to reduce that any further to determine what theta is...

Any help?

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- Jan 14th 2010, 08:49 AMpaupsers[SOLVED] Help determining principal Arg
Can anyone show me how to determine the Arg(2+i) ?

I know that $\displaystyle \theta=sin^{-1}(\frac{1}{\sqrt{5}})$

But I don't know how to reduce that any further to determine what theta is...

Any help? - Jan 14th 2010, 09:04 AMPlato
$\displaystyle \text{Arg}(x + yi) = \left\{ {\begin{array}{rl}

{\arctan \left( {\frac{y}

{x}} \right),} & {x > 0} \\

{\arctan \left( {\frac{y}

{x}} \right) + \pi ,} & {x < 0\,\& \,y > 0} \\

{\arctan \left( {\frac{y}

{x}} \right) - \pi ,} & {x < 0\,\& \,y < 0} \\

\end{array} } \right.$ - Jan 14th 2010, 11:03 AMpaupsers
Sorry, I should've put the whole problem on here. The problem states:

"By writing the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular coordinates, show that

$\displaystyle \frac{5i}{2+i}=1+2i$

This can be easily shown by just multiplying top and bottom by the conjugate of the denominator, but I have to do it how the directions say. I know how to convert 5i into exponential form, but converting 2+i into exponential form has proven to be more problematic since arctan(1/2) isn't very simple.

Any help? - Jan 14th 2010, 12:20 PMPlato
- Jan 14th 2010, 01:34 PMpaupsers
- Jan 14th 2010, 02:01 PMPlato
- Jan 14th 2010, 03:14 PMpaupsers
- Jan 14th 2010, 03:21 PMpaupsers
Nevermind! I figured it out!

Thanks!