Argument of a complex number

**craig** · January 16th 2009, 05:04 AM

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?

Thanks in advance

**mr fantastic** · January 16th 2009, 05:17 AM

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?

Thanks in advance

-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 ....

**craig** · January 16th 2009, 05:33 AM

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

**Plato** · January 16th 2009, 08:51 AM

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

**craig** · January 16th 2009, 08:55 AM

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

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