[SOLVED] Help determining principal Arg

**paupsers** · January 14th 2010, 08:49 AM

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?

**Plato** · January 14th 2010, 09:04 AM

$\text{Arg}(x + yi) = \left\{ {\begin{array}{rl} {\arctan \left( {\frac{y} {x}} \right),} & {x > 0} \\ {\arctan \left( {\frac{y} {x}} \right) + \pi ,} & {x < 0\,\& \,y > 0} \\ {\arctan \left( {\frac{y} {x}} \right) - \pi ,} & {x < 0\,\& \,y < 0} \\ \end{array} } \right.$

**paupsers** · January 14th 2010, 11:03 AM

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?

**Plato** · January 14th 2010, 12:20 PM

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.

**paupsers** · January 14th 2010, 01:34 PM

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.

**Plato** · January 14th 2010, 02:01 PM

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 $\frac{{5i}}{{2 + i}} = \sqrt 5 e^{\frac{i\pi}{2} } \left( {e^{ - i\phi } } \right) = \sqrt 5 e^{i\left( {\frac{\pi}{2} - \phi } \right)}$ .

**paupsers** · January 14th 2010, 03:14 PM

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.

**paupsers** · January 14th 2010, 03:21 PM

Nevermind! I figured it out!
Thanks!

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