# [SOLVED] Help determining principal Arg

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

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

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

Any help?
• Jan 14th 2010, 09:04 AM
Plato
$\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 AM
paupsers
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
$\frac{5i}{2+i}=1+2i$

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 PM
Plato
Quote:

Originally Posted by paupsers
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
$\frac{5i}{2+i}=1+2i$"

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.

You should always post the entire question.

Let $\phi = \arctan \left( {\frac{1}{2}} \right)$ then $
2 + i = \sqrt 5 e^{i\phi }$

Now do it.
• Jan 14th 2010, 01:34 PM
paupsers
Quote:

Originally Posted by Plato
You should always post the entire question.

There is a mistake in you question: it should be $\frac{5i}{2+i}=2{\color{red}-}i$
Let $\phi = \arctan \left( {\frac{1}{2}} \right)$ then $
2 + i = \sqrt 5 e^{i\phi }$

Now do it.

Actually, the way I posted the question is correct.
• Jan 14th 2010, 02:01 PM
Plato
Quote:

Originally Posted by paupsers
Actually, the way I posted the question is correct.

You are correct, I missed the 5i

Anyway,it still works.
$5i=5e^{\frac{i\pi}{2}}$ so $\frac{{5i}}{{2 + i}} = \sqrt 5 e^{\frac{i\pi}{2} } \left( {e^{ - i\phi } } \right) = \sqrt 5 e^{i\left( {\frac{\pi}{2} - \phi } \right)}$.
• Jan 14th 2010, 03:14 PM
paupsers
Quote:

Originally Posted by Plato
You are correct, I missed the 5i

Anyway,it still works.
$5i=5e^{\frac{i\pi}{2}}$ so $\frac{{5i}}{{2 + i}} = \sqrt 5 e^{\frac{i\pi}{2} } \left( {e^{ - i\phi } } \right) = \sqrt 5 e^{i\left( {\frac{\pi}{2} - \phi } \right)}$.

This is how far I had gotten before. This is where I don't understand how to get it back into rectangular coordinates.
• Jan 14th 2010, 03:21 PM
paupsers
Nevermind! I figured it out!
Thanks!