i got  : C

 

like this

 

let z = a+ib

    w = c+id

 

e^z = e^w

=>  e^(a+ib) = e^(c+id)

=>  e^(a-c) = e^i(d-b)

now LHS is real and RHS is complex and they r equal

it can only mean that  d - b = n(pie)

also  e^(a-c) = cos(d-b) = cos(npie) = -1 or 1  (depending on n )

now e^(any number ) is always positive

therefore e^(a-c) = 1 (which implies n is even)

=>  a - c = 0

 

now z - w  = (a-c) - i (d-b) =  -in(pie)

=>  Iz-wI = n(pie) (where n is even )

hence Iz-wI = 2(pie) or 4(pie) or 6(pie) ...................

 

hence it is C

 

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