I would need help with this question:
$$Z = \frac{(1+j2)^2(4-j3)^3 }{ (3+j4)^4 (2-j3)}$$
My starting point for this question is to expand the complex numbers first then continue doing but after expanding I am stuck.
Am I on the right track? Kindly help :D.
HINT:
For modulus use $\left|\dfrac{a^mb^n}{c^pd^q}\right|=\dfrac{|a|^m|b|^n}{|c|^p|d|^q}$ where $m,n,p,q$ are real
For the principal argument($\in(-\pi,\pi]$) (definition),
$$A=2\arctan\dfrac21+3\arctan\dfrac{-3}4-4\arctan\dfrac43-\arctan\dfrac{-3}2$$
using showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$ and Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
$2\arctan2=\pi+\arctan\dfrac{2\cdot2}{1-2^2}=\pi-\arctan\dfrac43$
$$\implies A=\pi-3\arctan\dfrac34-3\arctan\dfrac43+\arctan\dfrac32$$
$$=\pi-3\left(\text{arccot}\dfrac43+\arctan\dfrac43\right)+\arctan\dfrac32$$
$$=\pi-3\cdot\dfrac\pi2+\dfrac\pi2-\arctan\dfrac23$$
$$\equiv-\arctan\dfrac23\pmod{2\pi}$$
Since $-j(3+j4)=4-j3$, it reduces to $$Z=\frac{(1+j2)^2(4-j3)^3}{(3+j4)^4(2-j3)}=\frac{(1+j2)^2}{(2-j3)(3+j4)}\cdot\left(\frac{4-j3}{3+j4}\right)^3=\frac{(1+j2)^2}{(2-j3)(3+j4)}\cdot(-j)^3$$ Expanding numerator and denominator separately gives $$Z=\frac{-4-3j}{18-j}$$ Now multiplying it by $\frac{18+j}{18+j}$ gives $$Z=\frac{-4-3j}{18-j}\cdot\frac{18+j}{18+j}=\frac{-4\times 18+3+(-4-3\times 18)j}{18^2+1}=\frac{-69-58j}{325}$$ I'm sure you can continue from here.
Generally, after expanding the above you get something that looks like:
$$ Z =\frac{a+bj}{x+yj}$$
You can simplify this by rationalising the denominator:
$$Z= \frac{a+bj}{x+yj} \times\frac{x-yj}{x-yj} $$
This will now give you a complex number of the form: $p+qj$
Modulus(Z) $ = \sqrt{p^2 + q^2}$
$\arg(Z) = \arctan \left(\frac{q}{p}\right)$
There are some very good tricks in the other answers so you can take advantage of them.
