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I have been asked to solve the following and represent the answer graphically:

A) $| \arg z - (\pi/4) | < (\pi/2)$

I understand that this means the difference between the argument of $z$ and $(\pi/4)$ has to lie in the first quadrant, but I'm not quite certain how to represent this on a graph.

B) $| \arg ( 1+ i )z| \leq (\pi/2)$.

This one and the next are completely lost on me.

C) $| \arg z - \arg (1+i)| < (\pi/2)$.

I assume here $\arg (1+i)$ would be $\tan^{-}1 (1/1) = (\pi/4)$ and that would make it exactly like the first problem?

D) $|z-i|/|z+i| \leq 1$.

I know how to solve this one, but I'm not quite sure what area (above or below the perpendicular bisector of the line joining -i and i) I'm required to shade/mark. An explanation as to how I should figure this out would be appreciated!

Thanks a lot in advance for all your help!

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  • $\begingroup$ it would be useful for you to learn the basics of mathjax to improve the look of your question. e.g. if you enclose the expression "\pi" between two dollar signs you get $\pi$. right click on this latter. select "show math as" then "tex commands" $\endgroup$ Aug 20, 2014 at 11:00
  • $\begingroup$ Another point about this site: if you receive an answer you find useful, it is polite to accept one of the answers. This give the person some reward for helping you (and the first time gives you some too). I see that you have not yet done this on any question you have asked. $\endgroup$
    – Jessica B
    Aug 20, 2014 at 12:09
  • $\begingroup$ Sorry, I'm still relatively new to this website. $\endgroup$
    – user104389
    Aug 20, 2014 at 14:26

1 Answer 1

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Hint: One way to approach these sorts of questions is to try and think geometrically about what the equation is saying. For example, for D, rearranging gives $|z-i|\leq|z+i|$, which in words says '$z$ is closer to $i$ than to $-i$'.

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  • $\begingroup$ So, that would be above the bisector, if I've understood this correctly? $\endgroup$
    – user104389
    Aug 20, 2014 at 11:20
  • $\begingroup$ Yes, in this case. $\endgroup$
    – Jessica B
    Aug 20, 2014 at 11:22
  • $\begingroup$ Thanks, that was a great help! Do you think you could give me a hint for the others too, please? Those are more of a hassle for me at the moment. $\endgroup$
    – user104389
    Aug 20, 2014 at 11:25
  • $\begingroup$ For A, what values can $\arg z$ be to satisfy the equation? $\endgroup$
    – Jessica B
    Aug 20, 2014 at 11:30
  • $\begingroup$ Thanks, I figured it out! :) $\endgroup$
    – user104389
    Aug 20, 2014 at 14:26

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