Question 577718
Lets see if x = -2i is a solution. If it is indeed a true solution, then f(-2i) should equal 0 (ie f(-2i) = 0)



{{{f(x)=3x^3-x^2+4x-3}}} Start with the given equation



{{{f(-2i)=3(-2i)^3-(-2i)^2+4(-2i)-3}}} Plug in {{{x=-2i}}}



{{{f(-2i)=3(-8i^3)-4i^2-8i-3}}} Cube -2i to get {{{-8i^3}}}



{{{f(-2i)=-24i^3-4i^2-8i-3}}} Square -2i to get {{{4i^2}}}



{{{f(-2i)=-24(-i)-4i^2-8i-3}}} Replace {{{i^3}}} with -i (since {{{i^3=-i}}})



{{{f(-2i)=-24(-i)-4(-1)-8i-3}}} Replace {{{i^2}}} with -1 (since {{{i^2=-1}}})



{{{f(-2i)=24i+4-8i-3}}} Multiply



{{{f(-2i)=1+16i}}} Combine like terms



Since {{{f(-2i)=1+16i}}}, and f(-2i) is NOT equal to zero, this means that x = -2i is NOT a zero or a solution to the original equation.


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Now lets see if x = i is a solution. The same reasoning is used before and we're checking to see if f(i) = 0 



{{{f(x)=3x^3-x^2+4x-3}}} Start with the given equation



{{{f(i)=3i^3-i^2+4i-3}}} Plug in {{{x=i}}}



{{{f(i)=3(-i)-4i^2-8i-3}}} Cube i to get {{{-i}}} (since {{{i^3=-i}}})



{{{f(i)=3(-i)-4(-1)-8i-3}}} Square i to get -1 (since {{{i^2=-1}}})



{{{f(i)=-3i+4-8i-3}}} Multiply



{{{f(i)=1-11i}}} Combine like terms



Since {{{f(i)=1-11i}}}, and f(i) is NOT equal to zero as well, this means that x = i is NOT a zero or a solution to the original equation.




So neither x = -2i or x = i are solutions. 



So the answer is "none are zeros"