Complex though simple.

Question:
If !!(i(z^3) + Z^2 - z + i = 0)!!.
Then !!mod(z)!! !!=!! !!?!!
!!a)" "Rational!!

!!b)" "Integer!!

!!c)" "None!!

Solution !!-1 :!!

We can simplify this equation by simply taking !!i^2!! !!=!! !!-1!! before !!z!! and then factorizing the equation to find a root of this equation.
That is
!!(i(z^3) + Z^2 + i^2(z) + i)!! !!=!! !!0!!

!![z^2(1+iz) + i(1+iz)]!! !!=!! !!0!!

!!(z^2+i)(1+iz)!! !!=!! !!0!!

This gives
!!(z^2)!! !!=!! (-i) and !!z!! !!=!! !!(-i)!! and !!mod(z)!! !!=!! !!1!! which is an integer. So the option !!(b)!! is correct.

Solution!!-2:!!
We can put !!z!! !!=!! !!(e^(iθ))!! in the above equation and then factorize the equation to get the roots.
That is
!!(i(e^(iθ))^3 + (e^iθ)^2 - (e^iθ) + i)!! !!=!! !!0!!

Upon factoring we get;
!!(ie^(2iθ)-1)(e^(iθ)-i)!! !!=!! !!0!!

That is
!!z!! !!=!! !!i!! and !!z^2!! !!=!! !!i!! So !!Mod(z)!! !!=!! !!1!! which is an integer.
So option (b) is correct.

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