Proof Complex Numbers: How to Prove |z1|^2/|z2|^2 = |z1/z2|^2

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To prove the equation |z1|^2/|z2|^2 = |z1/z2|^2 for complex numbers, one can start by defining z1 and z2 in their standard forms, z1 = a + jb and z2 = l + jm. The left-hand side (LHS) involves calculating the modulus of each complex number, resulting in a^2 + b^2 for |z1|^2 and l^2 + m^2 for |z2|^2. For the right-hand side (RHS), the division of complex numbers can be rationalized by multiplying by the conjugate, leading to the expression (a + jb)(l - jm)/(l^2 + m^2). By separating the real and imaginary parts, it can be shown that the LHS equals the RHS, thus proving the equation. This method effectively demonstrates the relationship between the moduli of complex numbers and their division.
Suni
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hello

could someone please help me out with proving the following:
|z1|^2/|z2|^2 = |z1/z2|^2

...with complex numbers

sorry I am not familiar with the coding here yet so i can't write that properly
 
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How is division defined on the complex numbers?
Use that to get the result, alternatively use polar representation of the numbers.
 
simply define z1 = a + jb and z2 = l + jm;
LHS - take modulus of each and keep it aside...i.e., a^2 + b^2 is modulus;
RHS - Rationalise the den. & num., i.e, z1/z2 = (a + jb)*(l - jm)/(l2 + m2);
seperate out the real & imaginary parts...u'll find LHS = RHS...
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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