1. The problem is to solve the equation $$\left(z - \frac{1}{\sqrt{2}}\right)^{105} + \left(z + \frac{1}{\sqrt{2}}\right)^{105} = 0$$ where $z \in \mathbb{C}$. 2. This is a complex equation involving powers of binomials. We want to find all complex numbers $z$ satisfying it. 3. Rewrite the equation as: $$\left(z - \frac{1}{\sqrt{2}}\right)^{105} = -\left(z + \frac{1}{\sqrt{2}}\right)^{105}$$ 4. Taking the 105th root on both sides, we get: $$z - \frac{1}{\sqrt{2}} = \omega \left(z + \frac{1}{\sqrt{2}}\right)$$ where $\omega$ is a 105th root of $-1$. 5. The 105th roots of $-1$ are given by: $$\omega_k = e^{i\frac{\pi + 2\pi k}{105}}, \quad k=0,1,\ldots,104$$ 6. Substitute $\omega = \omega_k$ and solve for $z$: $$z - \frac{1}{\sqrt{2}} = \omega_k \left(z + \frac{1}{\sqrt{2}}\right)$$ 7. Rearranging: $$z - \omega_k z = \frac{1}{\sqrt{2}} (\omega_k + 1)$$ $$z(1 - \omega_k) = \frac{1}{\sqrt{2}} (\omega_k + 1)$$ 8. Provided $1 - \omega_k \neq 0$, solve for $z$: $$z = \frac{\frac{1}{\sqrt{2}} (\omega_k + 1)}{1 - \omega_k} = \frac{\omega_k + 1}{\sqrt{2}(1 - \omega_k)}$$ 9. Note that $1 - \omega_k = 0$ only if $\omega_k = 1$, but $\omega_k$ are 105th roots of $-1$, so $\omega_k \neq 1$. 10. Therefore, the solutions are: $$\boxed{z_k = \frac{\omega_k + 1}{\sqrt{2}(1 - \omega_k)}, \quad k=0,1,\ldots,104}$$ This gives 105 distinct complex solutions corresponding to the 105th roots of $-1$. Final answer: $$z_k = \frac{\omega_k + 1}{\sqrt{2}(1 - \omega_k)}$$ where $$\omega_k = e^{i\frac{\pi + 2\pi k}{105}}, k=0,1,\ldots,104.$$