You're Missing One Term

Hint: Consider the binomial expansion of $(1+z)^n$

Upon observation, the LHS of this complex-valued equation is just the Binomial Expansion of $(z + 1)^n$ sans the $z^n$ term. We have:

$(z+1)^n - z^n = 0 \Rightarrow (z+1)^n = z^n$ (i)

Taking $z = a + bi; a,b \in \mathbb{R}$ , we now transform (i) into a complex-exponential equivalent form:

$[(a+1) + bi]^n = (a+bi)^n$ ;

or $[(a+1)^2 + b^2]^{n/2} e^{in \cdot arctan(b/(a+1))} = (a^2 + b^2)^{n/2} e^{in \cdot arctan(b/a)};$ (ii).

Equating the moduli in (ii) yields $(a+1)^2 + b^2 = a^2 + b^2 \Rightarrow \boxed{a = -\frac{1}{2}}$ . Also, examination of the corresponding angles shows that:

$arctan(b/(a+1)) = arctan(b/a) \Rightarrow \frac{b}{a+1} = \frac{b}{a} \Rightarrow \frac{b}{1/2} = -\frac{b}{1/2}$

which means that the complex numbers $(a+1) + bi$ and $a + bi$ must form a complex-conjugate pair for $a = -\frac{1}{2}, b = K (K \in \mathbb{R})$ . Therefore, the roots are $\boxed{z = -\frac{1}{2} \pm Ki}$ which lie along a straight vertical line in the complex plane.

Come on this question doesn't need solution does it.!!!!

I wonder how is this on level 5.