# Complex numbers struggling

#### 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. $$0 < Arg\left( {z - 1 - i} \right) < \frac{\pi }{3}$$

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 $$0 < Arg\left( {\left( {x - 1} \right) + i\left( {y - 1} \right)} \right) < \frac{\pi }{3}$$.

Any help is appreciated.

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#### AKG

Homework Helper
Benny said:
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. $$0 < Arg\left( {z - 1 - i} \right) < \frac{\pi }{3}$$

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 $$0 < Arg\left( {\left( {x - 1} \right) + i\left( {y - 1} \right)} \right) < \frac{\pi }{3}$$.

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.

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