# Degree measure for an argument of imaginary numbers

• Jul 30th 2011, 05:24 PM
sterces
Degree measure for an argument of imaginary numbers
I take an online trig class and every now and again they like to throw in problems we've never seen before, this would be one of those days. An answer would be helpful obviously but if you could also explain how you got it would be great.

The argument of -3a + 4ai is_______ °. Round the answer to one decimal place.
• Jul 30th 2011, 05:53 PM
Plato
Re: Degree measure for an argument of imaginary numbers
Quote:

Originally Posted by sterces
The argument of -3a + 4ai is_______ °. Round the answer to one decimal place.

In this case I am breaking a long held principle on my part.
I am going to give you a complete solution.
WHY? Because I think still using degrees is stupid.
This answer is $\left[ {\pi - \arctan \left( {\frac{4}{3}} \right)} \right]^ \circ$
• Jul 31st 2011, 05:04 AM
mr fantastic
Re: Degree measure for an argument of imaginary numbers
Quote:

Originally Posted by Plato
In this case I am breaking a long held principle on my part.
I am going to give you a complete solution.
WHY? Because I think still using degrees is stupid.
This answer is $\left[ {\pi - \arctan \left( {\frac{4}{3}} \right)} \right]^ \circ$

I think your pi needs to be heated up to 180 degrees .....
• Jul 31st 2011, 05:11 AM
Plato
Re: Degree measure for an argument of imaginary numbers
Quote:

Originally Posted by mr fantastic
I think your pi needs to be heated up to 180 degrees .....

The degree operator is applied to entire expression in the brackets.
At least that is the it is done in most CAS.
So the $\pi$ does become $180^o$.
The reason for that is that $\arctan(4/3)$ is not in degrees.