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Elementary approach:

$\begin{aligned}\frac{(1+i)^{103}}{(1-i)^{101}}=\frac{(1+i)^{103}\cdot(1+i)^{101}}{\bigg((1-i)(1+i)\bigg)^{101}}&=\frac{\bigg((1+i)^2\bigg)^{102}}{(1+1)^{101}}\\&=\frac{(2i)^{102}}{2^{101}}=2\cdot i^{102}=2\cdot (-1)^{51}=(-2)\end{aligned}$

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Using Euler's Formula / De-Moivre's Theorem:

Expressing $z=(1+i)$ in polar form gives $z=re^{i\theta}$ with $r=\sqrt2$ and $\theta=\frac{\pi}{4}$ . Also, it gives us that $(1-i)=\overline{z}=re^{-i\theta}$

$\begin{aligned}\frac{(1+i)^{103}}{(1-i)^{101}}=\frac{(re^{i\theta})^{103}}{(re^{-i\theta})^{101}}&=r^2\cdot e^{204i\theta}\\&=(\sqrt2)^2\cdot e^{51\pi i}\\&=2\bigg(\cos(51\pi)+i\sin(51\pi)\bigg)=2(-1+0i)=(-2)\end{aligned}$

(1 + i)^2 = 2i

(1 + i)/(1 - i) = i

i^101 = i

The given expression = (2i)(i) = -2

$\frac { { (1+i) }^{ 103 } }{ { (1-i) }^{ 101 } }$ Evaluating the following : $(1+i) ^{ 2 } = 2i$ $(1-i) ^{ 2 } = -2i$ Therefore, the following equation $\frac { { (1+i) }^{ 100 } { (1+i) }^{ 3 } }{ { (1-i) }^{ 100 } { (1-i) } }$ simplifies to $\frac { { (1+i) }^{ 3 } }{ { (1-i) } }$ $\frac { { (1+i) }^{ 2 } { (1+i) } }{ { (1-i) } }$ Now, $\frac { { (1+i) } }{ { (1-i) } } = i$ Therefore we get, $(2i) (i)$ $\boxed{-2}$

$(1+i)^2=2i$

$\frac{1+i}{1-i}=i$

$\frac{(1+i)^{103}}{(1-i)^{101}}=(\frac{1+i}{1-i})^{101}(1+i)^2=i^{101}\times2i=\boxed{-2}$