Module and square of a complex number

Question

if I have the square module of a complex number, how can I solve it? Example:

$|e^{-1}\cos(2 \pi f t)-e^{-1}j\sin(2 \pi f t)-1|^2$

First idea: $[e^{-1}\cos(2 \pi f t)]^2 +[-e^{-1}\sin(2 \pi f t)-1] ^2$ theoretically $a^{2}+b^{2}+c^{2}+2bc$

Second idea: $[e^{-1}\cos(2 \pi f t) -e^{-1}\sin(2 \pi f t)-1] ^2$ theoretically $ a^{2}+b^{2}+c^{2}-2ab+2bc-2ac $

what is the correct solution? Thanks

Answer 1

$$|z|^2=\Re^2z+\Im^2z=(r\cos\theta-1)^2+(r\sin\theta)^2.$$

You can rewrite as

$$r^2-2r\cos\theta+1.$$

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Here it is not a very good idea to square before taking the modulus. If you want to, the correct way is

$$(a-ib+c)^2=a^2-b^2+c^2-2iab+2ac-2ibc$$

and

$$|a-ib+c|^2=\sqrt{(a^2-b^2+c^2+2ac)^2+(2ab+2bc)^2}.$$

If you aggreate $s:=a+c$,

$$\sqrt{(s^2-b^2)^2+(2sb)^2}=\sqrt{s^4+2s^2b^2+b^4}=s^2+b^2.$$