# Thread: Argument of a complex number

1. ## 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?

2. 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 ....

3. 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

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

5. 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