How Complex Is It?

Relevant wiki: De Moivre's Theorem

$\begin{aligned} LHS & = (1+i)(-1+i)(-1-i)(1-i) \\ & = \sqrt 2 \left(\frac 1{\sqrt 2} +\frac i{\sqrt 2}\right) \cdot \sqrt 2 \left(-\frac 1{\sqrt 2} +\frac i{\sqrt 2}\right) \cdot \sqrt 2 \left(-\frac 1{\sqrt 2} - \frac i{\sqrt 2}\right) \cdot \sqrt 2 \left(\frac 1{\sqrt 2} - \frac i{\sqrt 2}\right) \\ & = \sqrt 2 e^{\frac \pi 4 i} \cdot \sqrt 2 e^{\frac {3\pi} 4 i} \cdot \sqrt 2 e^{\frac {5\pi} 4 i} \cdot \sqrt 2 e^{\frac {7\pi} 4 i} \\ & = 4 e^{4 \pi i} = 4 \end{aligned}$

$\begin{aligned} RHS & = \frac 14 (1+\sqrt 3i)(-1+\sqrt 3i)(-1-\sqrt 3i)(1-\sqrt 3i) \\ & = \frac 14 \cdot 2 \left(\frac 12 +\frac {\sqrt 3}2i \right) \cdot 2 \left(-\frac 12 +\frac {\sqrt 3}2i \right) \cdot 2 \left(-\frac 12 -\frac {\sqrt 3}2i \right) \cdot 2 \left(\frac 12 - \frac {\sqrt 3}2i \right) \\ & = \frac {16}4 e^{\frac \pi 3 i} \cdot e^{\frac {2\pi} 3 i} \cdot e^{\frac {4\pi} 3 i} \cdot e^{\frac {5\pi} 3 i} \\ & = 4 e^{4 \pi i} = 4 \end{aligned}$

$\implies LHS = RHS$ , therefore, the answer is $\boxed{\text{True}}$