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 February 22nd, 2014, 08:16 PM #1 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Find the image Find the image of the ray $arg(z)=\pi/3$ under $f(z)=2z^2 + 1 - i$ The answer I came up with on my own is a ray w/ $\theta= 2\pi/3$ emitting from 1-i. The answer in the book has two things I do not understand. States that $1-i$ is not in the image and that $(\sqrt{3}-1)i$ is contained in the image. Thanks for any help on this.
 February 23rd, 2014, 02:02 AM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Find the image Work with polar coordinates again.
 February 23rd, 2014, 08:59 AM #3 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Re: Find the image start with $\frac{\pi}{3}$ ; $2z^2 + 1 - i$ a)$r^2 2e^{i\frac{2\pi}{3}} + 1 - i$ b)$r^2 2(-.5 + \frac{\sqrt{3}}{2}i) + 1 - i$ c)for r = 1 $-1 + \sqrt{3}i + 1 -i$ d)$(sqrt{3}-1)i$ So far, am I approaching this correctly? Still not seeing how it is clear that $1-i$ is not in the image.
 February 24th, 2014, 05:24 AM #4 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Find the image Ok, your answer is correct and it describes the ray emitting from $1-i$ using function compositions. But why $(1-i)$ is not contained in the image this is simply because $z=0$ is not contained in $arg(z)= \frac{\pi}{3}$. The definition of argument states that $arg(0)$ is undefined since polar coordinates are not useful to describe the point $z=0$. If we say that the point $z=0$ describes the vector of radius $r=0$ then clearly the direction cannot be specified because a point has no direction. So $z=0$ is not mapped by the function $f(z)= z^2+1-i$ so $f \neq 1-i$ for otherwise $z=0$.

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