My Math Forum Geometric Relationship

 Complex Analysis Complex Analysis Math Forum

 February 22nd, 2014, 11:41 AM #1 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Geometric Relationship The question asks how to describe, geometrically, the relationship between a nonzero complex number, $z=a+ib$, and its inverse, $z^{-1}$ I see that $z^{-1}=\frac{a-ib}{a^2+b^2}$ for $a$ or $b$ $>0$ shrinks, and reflects each number about either axis by plugging in a few values for each. I am having a hard time coming up w/ a concise way of describing this and also wondering it there is another way of looking at this that makes this geometric relationship more obvious. Thanks for any help!
 February 22nd, 2014, 11:44 AM #2 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Geometric Relationship Ok, scratch reflects each number about either axis..... $z$ and $z^{-1}$ are symmetric about the origin.
 February 22nd, 2014, 01:05 PM #3 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Geometric Relationship Hint : use that $z= re^{i \theta }$
 February 22nd, 2014, 09:31 PM #4 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Geometric Relationship Now I have more questions, lol How do I apply the inverse function to $z=re^{i\theta}$? $z^{-1}= \frac {1}{r}ln(i\theta)$?
 February 23rd, 2014, 02:00 AM #5 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Geometric Relationship Let $w=\frac{1}{z}$ then the mapping of the point $z=r e^{i \theta }$ under $w$ is what ?
 February 23rd, 2014, 08:22 AM #6 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Geometric Relationship I see now.... I think! $w=\frac{1}{r}e^{-i\theta}$
February 23rd, 2014, 09:54 AM   #7
Math Team

Joined: Aug 2012
From: Sana'a , Yemen

Posts: 1,177
Thanks: 44

Math Focus: Theory of analytic functions
Re: Geometric Relationship

Quote:
 Originally Posted by WWRtelescoping I see now.... I think! $w=\frac{1}{r}e^{-i\theta}$
Correct , can you see know what happen to the modulus and the argument ?

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