# Complex numbers: Argument of Z help

• August 6th 2009, 04:27 AM
RAz
Complex numbers: Argument of Z help
I'm having some problems with finding, in radians, the argument of z and adding rotations of 180 to keep it in range. I know that I have to use:

$tan=\frac{y}{x}$

and I know exact trigonometric ratios, and the ASTC in the polar plane. But I don't know where Arg Z is positive, and I don't know when to add $2\pi$ or even just $\pi$ (pi = 180 in radians), to keep Arg Z in the range of $-\pi < 0 <\pi$. As you can see I am very confused, any help would be great.
• August 6th 2009, 05:21 AM
Calculus26
Consider 1 + i and - 1 - i

In both cases arctan(y/x) = pi/4

since 1 + i is in the first quadr arg(z) = pi/4

hoewever for - 1 -i we know this is in the 3d quadrant so we add pi for the arg

Similarly consider 1 - i and -1 + i

arctan(y/x) = -pi/4 for -1 + i is in the 2 quadrant so add pi to find argz
for 1- i is in the fourth quadrant so add 2pi to find argz

In general locate the quadrant in which z lies to determine arg z
• August 6th 2009, 05:22 AM
mr fantastic
Quote:

Originally Posted by RAz
I'm having some problems with finding, in radians, the argument of z and adding rotations of 180 to keep it in range. I know that I have to use:

$tan=\frac{y}{x}$

and I know exact trigonometric ratios, and the ASTC in the polar plane. But I don't know where Arg Z is positive, and I don't know when to add $2\pi$ or even just $\pi$ (pi = 180 in radians), to keep Arg Z in the range of $-\pi < 0 <\pi$. As you can see I am very confused, any help would be great.

Plot your value of z on an Argand diagram. That makes it very easy to see the ballpark your argument lies in.
• August 6th 2009, 06:07 AM
Plato
This function will work.
$
\arg \left( {x + yi} \right) = \left\{ {\begin{array}{ll}
{\arctan \left( {y/x} \right),} & {x > 0} \\
{\arctan \left( {y/x} \right) + \pi ,} & {x < 0\;\& \,y > 0} \\
{\arctan \left( {y/x} \right) - \pi ,} & {x < 0\;\& \,y < 0} \\

\end{array} } \right.$
• August 6th 2009, 04:06 PM
HallsofIvy
Since you want to keep Arg z between $-\pi$ and $\pi$, Arg z is positive as long as the complex part is postive, negative if the comples part is negative.
• August 6th 2009, 04:45 PM
pankaj
Quote:

Originally Posted by RAz
I'm having some problems with finding, in radians, the argument of z and adding rotations of 180 to keep it in range. I know that I have to use:

$tan=\frac{y}{x}$

and I know exact trigonometric ratios, and the ASTC in the polar plane. But I don't know where Arg Z is positive, and I don't know when to add $2\pi$ or even just $\pi$ (pi = 180 in radians), to keep Arg Z in the range of $-\pi < 0 <\pi$. As you can see I am very confused, any help would be great.

First find angle $\alpha$ using $\tan\alpha=\left|\frac{y}{x}\right|$,where $\alpha$ is an angle lying between $0$ and $\frac{\pi}{2}$

Now $arg(z)$ is defined as the angle which the line segment joining origin and $z$ makes with positive direction of the real axis.
Accordingly,
If $x>0,y>0$,then $arg(z)=\alpha$

If $x<0,y>0$,then $arg(z)=\pi-\alpha$

If $x<0,y<0$,then $arg(z)=\alpha-\pi$

If $x>0,y<0$,then $arg(z)=-\alpha$

If $y=0$ and $x>0$,then $arg(z)=0$,If $y=0$ and $x<0$,then $arg(z)=\pi$

If $x=0$ and $y>0$,then $arg(z)=\frac{\pi}{2}$,If $x=0$ and $y<0$,then $arg(z)=-\frac{\pi}{2}$

Draw all these cases on the graph and keep the definition of arg(z)(underlined part) foremost in your mind and you will understand immediately.
• August 6th 2009, 04:54 PM
Plato
Quote:

Originally Posted by Plato
This function will work.
$
\arg \left( {x + yi} \right) = \left\{ {\begin{array}{ll}
{\arctan \left( {y/x} \right),} & {x > 0} \\
{\arctan \left( {y/x} \right) + \pi ,} & {x < 0\;\& \,y > 0} \\
{\arctan \left( {y/x} \right) - \pi ,} & {x < 0\;\& \,y < 0} \\
\end{array} } \right.$

To the former two posters,
With the exception $x=0$ that function answers all of your points.
What problem do you have with that?
• August 6th 2009, 05:34 PM
pankaj
None whatsoever Plato.

It is just that I remembered the time when I had got stuck in the same concept and finally came out of it when my teacher had explained it to me in the manner I have written.
• August 6th 2009, 10:57 PM
RAz
Thanks for your help everyone. :)