Complex number geometrical problem

In summary: Let's take a look at a picture:What can we say about the angle (with the real axis) of $z$ in relation to the angle of $z+1$?
  • #1
amr21
11
0
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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? :)
 
  • #4
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 $z+1$?
 
  • #5
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 ?
 

FAQ: Complex number geometrical problem

1. What is a complex number?

A complex number is a number that has both a real part and an imaginary part. It is written in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).

2. How are complex numbers represented on a graph?

Complex numbers can be represented on a graph using a coordinate system called the complex plane, where the real part is on the x-axis and the imaginary part is on the y-axis. The complex number a+bi is plotted as the point (a,b) on the complex plane.

3. What is the geometric interpretation of multiplying complex numbers?

Multiplying two complex numbers is equivalent to scaling and rotating the complex plane. The magnitude of the product is the product of the magnitudes of the two complex numbers, and the angle of the product is the sum of the angles of the two complex numbers.

4. How do you find the conjugate of a complex number?

The conjugate of a complex number a+bi is a-bi. This means that the sign of the imaginary part is changed to its opposite, but the real part remains the same. Geometrically, the conjugate of a complex number is the reflection of the number across the real axis on the complex plane.

5. What is the modulus of a complex number?

The modulus, or absolute value, of a complex number a+bi is the distance from the origin (0,0) to the point (a,b) on the complex plane. It is calculated as √(a^2 + b^2). The modulus represents the magnitude or size of the complex number.

Similar threads

Replies
4
Views
1K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
2
Views
1K
Replies
2
Views
1K
Back
Top