<s>[INST] Find the absolute value and argument for $(3+4i)^{-1}$ When I have to find the argument and absolute value of $z=(3+4i)^{-1}$ I thought that I had to find the inverse of $z$ by applying the rule $z^{-1}=\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right)$ but the answer suggests that $\lvert(3+4i)^{-1}\rvert$=$\frac{1}{\lvert3+4i\rvert}$ and that $(Arg)z^{-1}=-Argz$
Why does my first approach give different solutions? Is it that the inverse definition of a complex number is not the same with the annotation $z^{-1}$?
 [/INST] The inverse of a complex number $z=a+bi$ is given by:
$$z^{-1} = \frac{1}{a+bi} = \frac{a-bi}{(a+bi)(a-bi)} = \frac{a-bi}{a^2+b^2}.$$
Thus, its absolute value is
$$|z^{-1}| = \sqrt{\frac{a^2}{(a^2+b^2)^2} + \frac{b^2}{(a^2+b^2)^2}} = \frac{\sqrt{a^2+b^2}}{a^2+b^2} = \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{|z|}.$$
Using the exponential, you can represent $z$ as
$$z=|z|e^{i\arg(z)}.$$
Thus
$$z^{-1}=|z^{-1}|e^{i\arg(z^{-1})} = \frac{1}{|z|} e^{i\arg(z^{-1})},$$
so
$$1 = z\,z^{-1} = |z|e^{i\arg(z)} \frac{1}{|z|} e^{i\arg(z^{-1})} = e^{i(\arg(z)+\arg(z^{-1}))} \Rightarrow \arg(z)+\arg(z^{-1}) = 0$$
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