Find the modulus and argument of a complex number

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To find the modulus and argument of the complex number z=((1+2i)^2 * (4-3i)^3) / ((3+4i)^4 * (2-i)^3), one can either multiply the factors in the numerator and denominator directly or convert each complex number to polar form for easier calculations. The modulus can be calculated using the formula mod(z)=sqrt(a^2+b^2), which requires correctly identifying the real and imaginary parts. It's important to ensure that the squares of both the real and imaginary parts are added, rather than subtracted, especially in the denominator. Properly applying these methods will lead to the correct modulus and argument.
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Homework Statement


Find the modulus and argument of

z=((1+2i)^2 * (4-3i)^3) / ((3+4i)^4 * (2-i)^3

Homework Equations


mod(z)=sqrt(a^2+b^2)

The Attempt at a Solution


In order to find the modulus, I have to use the formula below. But I'm struggling with finding out how to put the equation in the formula:
I have attached a photo of how I did it so far. But unfortunately, it does not give sence.
 

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javii said:

Homework Statement


Find the modulus and argument of

z=((1+2i)^2 * (4-3i)^3) / ((3+4i)^4 * (2-i)^3

Homework Equations


mod(z)=sqrt(a^2+b^2)

The Attempt at a Solution


In order to find the modulus, I have to use the formula below. But I'm struggling with finding out how to put the equation in the formula:
I have attached a photo of how I did it so far. But unfortunately, it does not give sence.
You could just multiply all of the factors in the numerator, and the factors in the denominator, and then do the division.
OR, you could rewrite each complex number in polar form and then do the multiplications and division.
 
javii said:

Homework Equations


mod(z)=sqrt(a^2+b^2)

To calculate the modulus, you have to add the square of both the real and imaginary parts. But you subtracted them in the denominator.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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