MHB Complex number geometrical problem

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The discussion centers on demonstrating geometrically that if |z|=1, then the imaginary part of the expression z/(z+1)^2 equals zero. Participants suggest focusing on the arguments of complex numbers, noting that the imaginary part is zero when the argument is either 0 or π. A geometric approach involves sketching the triangle formed by the points (0, z, z+1) and analyzing the angles. Visual aids, such as diagrams, are recommended to clarify the relationship between the angles of z and z+1. Understanding the conversion to arguments is emphasized as crucial for solving the problem.
amr21
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Show geometrically that if |z|=1 then, $Im[z/(z+1)^2]=0$

I am unsure how to begin this problem. I have sketched out |z|=1 but can't work out how to sketch the Imaginary part of the question.
 
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amr21 said:
Show geometrically that if |z|=1 then, $Im[z/(z+1)^2]=0$

I am unsure how to begin this problem. I have sketched out |z|=1 but can't work out how to sketch the Imaginary part of the question.

Hi amr21,

You should think in terms of arguments. In particular, for any complex numbers $a$ and $b$:

  • $\mbox{Im}(a) = 0$ if $\arg(a)=0$ or $\arg(a)=\pi$.
  • $\arg(ab) = \arg(a) + \arg(b) \pmod{2\pi}$

Try to use that to convert the question into a question about arguments.

You should then make a complete drawing. Pick a point $z$ such that $|z|=1$ and look at the triangle $(0,z,z+1)$.

Can you try that ? Write back if you need further help.
 
castor28 said:
Hi amr21,

You should think in terms of arguments. In particular, for any complex numbers $a$ and $b$:

  • $\mbox{Im}(a) = 0$ if $\arg(a)=0$ or $\arg(a)=\pi$.
  • $\arg(ab) = \arg(a) + \arg(b) \pmod{2\pi}$

Try to use that to convert the question into a question about arguments.

You should then make a complete drawing. Pick a point $z$ such that $|z|=1$ and look at the triangle $(0,z,z+1)$.

Can you try that ? Write back if you need further help.

I don't think I understand, could you explain further? :)
 
amr21 said:
I don't think I understand, could you explain further? :)

Let's take a look at a picture:
\begin{tikzpicture}[scale=3]
\coordinate (O) at (0,0);
\coordinate (Z) at ({sqrt(1/2)},{sqrt(1/2)});
\coordinate (Zp1) at ({sqrt(1/2)+1},{sqrt(1/2)});
\draw circle (1);
\draw (O) -- (Z) -- (Zp1) -- cycle;
\draw (O) -- (1,0) -- (Zp1);
\draw (1,0) -- (Z);
\path (O) node[below left] {0} -- (Z) node[above] {z} -- (Zp1) node[above right] {z+1} -- (1,0) node[below right] {1};
\path (O) -- node[below] {1} (1,0) -- node[below] {1} (Zp1) -- node[above] {1} (Z) -- node[above] {1} (O);
\end{tikzpicture}

What can we say about the angle (with the real axis) of $z$ in relation to the angle of $z+1$?
 
amr21 said:
I don't think I understand, could you explain further? :)
Hi amr21,

Could you also confirm that you understand the first part, about converting the question to a question about arguments ? If you don't, could you pinpoint what you don't understand ?
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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