Understanding Argand diagrams for complex numbers

[Benny]

Hi I'm struggling with the following questions where I need to sketch Argand diagrams. I haven't had much exposure to a wide range of these sortsof questions before so I'm not finding the following to be all that easy. There are a couple and some help would be good, thanks.

1. |z| < Argz.

Would this look like a spiral of increasing 'radius.' Like a swirly shape starting at the origin? Would the origin be included? I ask this because I don't think I can have |z| < 0.

Note: -pi < Argz <= pi.

2. log|z| = -2Argz.

Would I just exponentiate both sides to get \left| z \right| = e^{ - 2Argz} ?

If that's correct then what would the shape look like? Perhaps a 'circle' with a a varying radius?

3. 0 &lt; Arg\left( {z - 1 - i} \right) &lt; \frac{\pi }{3}

I don't know how to work with this one. The part, z - 1 - i just means the difference between z and (1+i) I think. Let z = x + yi so 0 &lt; Arg\left( {\left( {x - 1} \right) + i\left( {y - 1} \right)} \right) &lt; \frac{\pi }{3}.

Any help is appreciated.

 

[AKG]

1. |z| < Argz.

This is kind of a strange question, because you say Arg(z) may be negative, and |z| < Arg(z), but |z| > 0. Doing this for the range 0 to pi, thenyes, you'll see a spiral of increasing radius. Note that |z| < Arg(z), not just |z| = Arg(z), so your spiral should be "coloured in."

2. log|z| = -2Argz.

Would I just exponentiate both sides to get \left| z \right| = e^{ - 2Argz} ?

That's the right idea. For this type of problem (and the previous), just go through a few values of Arg(z), and figure out |z|, and plot your points. There's not really much room for confusion.

3. 0 &lt; Arg\left( {z - 1 - i} \right) &lt; \frac{\pi }{3}

I don't know how to work with this one. The part, z - 1 - i just means the difference between z and (1+i) I think. Let z = x + yi so 0 &lt; Arg\left( {\left( {x - 1} \right) + i\left( {y - 1} \right)} \right) &lt; \frac{\pi }{3}.

Any help is appreciated.

If you set z = x + iy, try finding the boundaries. So find Arg(z - (1 + i)) = 0. You know that this will have to be on the real axis, so y = 1, and x will range from -1 up to infinity, so you'll get the ray terminating at (-1, 1) and extending towards the right. Do the same for Arg(z - (1+i)) = pi/3, and for some intermediate angles as well for good measure. My guess at first glance that it will look like something like a Chinese paper fan or the Shell gas shell with it's "center" or pivot (if you think about the fan) at (1,1).

 

[Benny]

Thanks for your help.