# Argument of a complex number

• January 16th 2009, 05:04 AM
craig
Argument of a complex number
Hi, sorry another problem with the argument of a complex number :S.

1. Calculate $arg z$, giving your answer in radians to 2dp, where $z = -1 + 2i$.

Got the answer -1.07, which is right as $-\pi\leq-1.07\leq\pi$, yet the answers have the answer as $\pi + -1.07$.

Is this something I am missing, or just an odd answer?

• January 16th 2009, 05:17 AM
mr fantastic
Quote:

Originally Posted by craig
Hi, sorry another problem with the argument of a complex number :S.

1. Calculate $arg z$, giving your answer in radians to 2dp, where $z = -1 + 2i$.

Got the answer -1.07, which is right as $-\pi\leq-1.07\leq\pi$, yet the answers have the answer as $\pi + -1.07$.

Is this something I am missing, or just an odd answer?

-1.07 is in the 4th quadrant. -1 + 2i is in the 2nd quadrant.

You should always draw an argand diagram so you can see what quadrant z is in ....
• January 16th 2009, 05:33 AM
craig
Quote:

Originally Posted by mr fantastic
-1.07 is in the 4th quadrant. -1 + 2i is in the 2nd quadrant.

You should always draw an argand diagram so you can see what quadrant z is in ....

Thank you :)
• January 16th 2009, 08:51 AM
Plato
Craig, here is a simple scheme for doing the argument.
$\arg (z) = \arg (x + yi) = \left\{ {\begin{array}{ll}
{\arctan (y/x)} & {x > 0} \\
{\arctan (y/x) + \pi } & {x < 0\,\& \,y > 0} \\
{\arctan (y/x) - \pi } & {x < 0\,\& \,y < 0} \\ \end{array} } \right.$
• January 16th 2009, 08:55 AM
craig
Quote:

Originally Posted by Plato
Craig, here is a simple scheme for doing the argument.
$\arg (z) = \arg (x + yi) = \left\{ {\begin{array}{ll}
{\arctan (y/x)} & {x > 0} \\
{\arctan (y/x) + \pi } & {x < 0\,\& \,y > 0} \\
{\arctan (y/x) - \pi } & {x < 0\,\& \,y < 0} \\ \end{array} } \right.$

Thanks again :) Hopefully with this and drawing the diagram I should be ok ;)