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 Post subject: Complex Analysis - Question about methodology for a mapping.
PostPosted: Tue, 13 Mar 2012 08:47:46 UTC 
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Joined: Thu, 1 Mar 2012 10:14:08 UTC
Posts: 7
Show that the image of the open disk |z + 1 + i | < 1 under the transformation w = (3 - 4i)z + 6 + 2i is the open disk |w + 1 - 3i| < 5.

The method done by the book is as follows, through an inverse transformation:

z = (w - 6 - 2i)/(3 - 4i)

This is substituted into the first given inequality.

Simplifying down to |w - 6 - 2i + (1 + i)(3 - 4i)| < 5, which now goes to |w + 1 - 3i| < 5.

My method was to use the transformation w = .... directly and substitute that into |w + 1 - 3i| < 5

Doing it this way and simplifying it, we're left with the initial inequality |z + 1 + i| < 1, which means it's correct, but is the method right?

I do not fully understand this inverse transformation idea. If we have something (a point/a plane/whatever) in the z-plane, and want to show the corresponding image of it in the w-plane, why are we using an inverse transformation z = ... ?

I cannot get my head around this. Shouldn't we just the actual mapping function w = ... to prove that the image is the correct one?


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 Post subject: Re: Complex Analysis - Question about methodology for a mapp
PostPosted: Tue, 13 Mar 2012 15:25:07 UTC 
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Cryptid wrote:
Show that the image of the open disk |z + 1 + i | < 1 under the transformation w = (3 - 4i)z + 6 + 2i is the open disk |w + 1 - 3i| < 5.

The method done by the book is as follows, through an inverse transformation:

z = (w - 6 - 2i)/(3 - 4i)

This is substituted into the first given inequality.

Simplifying down to |w - 6 - 2i + (1 + i)(3 - 4i)| < 5, which now goes to |w + 1 - 3i| < 5.

My method was to use the transformation w = .... directly and substitute that into |w + 1 - 3i| < 5

Doing it this way and simplifying it, we're left with the initial inequality |z + 1 + i| < 1, which means it's correct, but is the method right?

I do not fully understand this inverse transformation idea. If we have something (a point/a plane/whatever) in the z-plane, and want to show the corresponding image of it in the w-plane, why are we using an inverse transformation z = ... ?

I cannot get my head around this. Shouldn't we just the actual mapping function w = ... to prove that the image is the correct one?


Unfortunately not, seeing that you did not mention the magic word.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Complex Analysis - Question about methodology for a mapp
PostPosted: Tue, 13 Mar 2012 20:05:19 UTC 
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Joined: Thu, 1 Mar 2012 10:14:08 UTC
Posts: 7
Sorry, what word is that?


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 Post subject: Re: Complex Analysis - Question about methodology for a mapp
PostPosted: Sat, 17 Mar 2012 08:33:50 UTC 
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Joined: Thu, 1 Mar 2012 10:14:08 UTC
Posts: 7
Anyone have an idea regarding this?


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