The duality property is the statement that if
$$\mathcal{F}(h(t)) = H(\omega)$$
then
$$\mathcal{F}(H(t)) = h(-\omega)$$
To use the duality property to prove the statement you need to show that
$$\mathcal{F}\left(i~\text{sign}(t)\right) = -\frac{1}{\omega\pi}$$
This can be done by a direct computation of $\mathcal{F}\left(i~\text{sign}(t)\right)$. To do this note that
$$\frac{d}{dx} \text{sign} (x) = 2\delta(x)$$
and by using the property
$$ \mathcal{F}\left(\frac{dg(x)}{dx}\right) = 2\pi i\omega \mathcal{F}(g(x))$$
togeather with $\mathcal{F}(\delta) = 1$ the result follows (I have used the convention $\mathcal{F}(g) \equiv \int g(x)e^{-2\pi i \omega} dx$ above).
For the sake of simplicity, let $t\in\mathbb{C}\setminus\mathbb{R}_{\geq 0}$ so that the polynomial $X^2-t\in\mathbb{C}[X]$ does not have a real root, but you can use the Cauchy principal values when $t\in\mathbb{R}_{\geq 0}$ to deal with the job. This job is to compute the Fourier transform $F_t$ of the function $f_t$, where $$f_t(z):=\dfrac{1}{z^2-t}\text{ for all }z\in\mathbb{C}\setminus\big\{-\sqrt{t},+\sqrt{t}\big\}\,,$$
where $\sqrt{t}$ is so chosen that the imaginary part is positive (this is possible as $X^2-t$ has no real roots).
That is,
$$F_t(k)=\frac{1}{\sqrt{2\pi}}\,\int_{-\infty}^{+\infty}\,\exp(\text{i}kx)\,f_t(x)\,\text{d}x\text{ for }k\in\mathbb{R}\,.$$
For simplicity, I shall write $c:=\dfrac{1}{\sqrt{2\pi}}$ (since $c$ may be defined to be a different constant for a different user, I shall leave this constant as just $c$).
Without loss of generality, suppose that $k\geq 0$. For $R>0$, let $C_R$ to be the positively oriented contour given by
$$[-R,+R]\cup\Big\{R\,\exp(\text{i}\theta)\,\Big|\,\theta\in[0,\pi]\Big\}\,.$$
Observe that
$$2\pi\text{i}\,\text{Res}_{z=+\sqrt{t}}\big(f_t(z)\big)=\lim_{R\to\infty}\,\oint_{C_R}\,\frac{\exp(\text{i}kz)}{z^2-t}\,\text{d}z=\frac{1}{c}\,F_t(k)\,.$$
Hence, taking the sign of $k$ into account, we get
$$F_t(k)=\frac{c\pi\text{i}}{\sqrt{t}}\,\exp\Big(\text{i}|k|\sqrt{t}\Big)\text{ for each }k\in\mathbb{R}\,.$$
For example, $\sqrt{+\text{i}}=\frac{+1+\text{i}}{\sqrt{2}}$ and $\sqrt{-\text{i}}=\frac{-1+\text{i}}{\sqrt{2}}$. This means
$$F_{s\text{i}}(k)=\frac{c\pi\text{i}}{\left(\frac{s+\text{i}}{\sqrt{2}}\right)}\,\exp\Biggl(\text{i}\,\left(\frac{s+\text{i}}{\sqrt{2}}\right)\,|k|\Biggr)=c\pi\left(\frac{1+s\text{i}}{\sqrt{2}}\right)\,\exp\left(-\frac{|k|}{\sqrt{2}}\right)\,\exp\left(\frac{s\text{i}|k|}{\sqrt{2}}\right)\,,$$
where $s\in\{-1,+1\}$. Now, let $\varphi(z):=\dfrac{1}{z^4+1}$. Then, as you found out,
$$\varphi(z)=-\frac{\text{i}}{2}\,\sum_{s\in\{-1,+1\}}\,\frac{s}{z^2-s\text{i}}\,.$$
Hence, the Fourier transform $\hat{\varphi}$ of $\varphi$ is given by
$$\hat{\varphi}(k)=-\frac{\text{i}}{2}\,\sum_{s\in\{-1,+1\}}\,s\,F_{s\text{i}}(k)\,.$$
That is,
$$\hat{\varphi}(k)=\frac{c\pi}{\sqrt{2}}\,\sum_{s\in\{-1,+1\}}\,\left(\frac{1-s\text{i}}{2}\right)\,\exp\left(-\frac{|k|}{\sqrt{2}}\right)\,\exp\left(\frac{s\text{i}|k|}{\sqrt{2}}\right)\,.$$
That is,
$$\hat{\varphi}(k)=\frac{c\pi}{\sqrt{2}}\,\exp\left(-\frac{|k|}{\sqrt{2}}\right)\,\Biggl(\cos\left(\frac{|k|}{\sqrt{2}}\right)+\sin\left(\frac{|k|}{\sqrt{2}}\right)\Biggr)\,.$$
Now, if you want to determine the Fourier transform $\hat{\Phi}$ of $\Phi$, where $$\Phi(z):=\dfrac{z^4}{z^4+1}=1-\dfrac{1}{z^4+1}\,,$$ you will get a distribution. Note that the Fourier transformation $\hat{\kappa}_\epsilon$ of the constant function $\kappa_\epsilon\equiv \epsilon$ is $$\hat{\kappa}_{\epsilon}(k)=2c\pi\,\epsilon\,\delta(k)\,,$$
where $\delta$ is the Dirac delta distribution.
Thus, using $\Phi=\kappa_1-\phi$, we conclude that
$$\hat{\Phi}(k)=2c\pi\,\left(\delta(k)-\frac{1}{2\sqrt{2}}\,\exp\left(-\frac{|k|}{\sqrt{2}}\right)\,\Biggl(\cos\left(\frac{|k|}{\sqrt{2}}\right)+\sin\left(\frac{|k|}{\sqrt{2}}\right)\Bigg)\right)\,.$$
Nonetheless, be warned that $\hat{\Phi}$ is not a function.
Best Answer
For the the second-to-last step, he's using the property that F[xf] = (F[f])'/i; that is, multiplying by x and taking the Fourier Transform is the same as taking the Fourier Transform, taking the derivative of that, and then dividing by i (or, equivalently, multiplying by -i). For the last step, he's simply using the definition of the Fourier Transform: to take the Transform, you integrate $e^{ikx}f(x)$. So to get the Transform of $e^{5ix}f(x)$, you integrate $e^{ikx}e^{5ix}f(x)=e^{ikx+5ix}f(x)=e^{i(k+5)x}f(x)$. If we define $u=k+5$, then this is just the Fourier Transform of $f(x)$, except that the transformed function is a function of $u$. To get back to $k$, we subtract 5. So your professor just replaced all the instances of $k$ in the the second-to-last step with $k-5$. That is, the "k" in the second-to-last step is 5 less than the "k" in the last step.
If you're having trouble with that:
Take
$\int e^{i(t+5)x}f(x)$
set $u = t+5$
$\int e^{iux}f(x)=F[f](u)$
since $u=t+5$, $t=u-5$, so $F[f](u)=F[f](t-5)$
So we take the Fourier Transform of $f$, then replace all instances of $k$ with $k-5$.
Remember, the Fourier Transform $F[f](k)$ is simply the strength of the $e^{ikx}$ component of $f$. By multiplying $f$ by $e^{5ix}$, we're shifting all the components by 5: the $k=0$ component becomes the $k=5$ component, etc.
There's a dual aspect to the translation property of Fourier Transforms: translating $f$ by $h$ multiplies the transform by $e^{ihx}$, and multiplying $f$ by $e^{ihx}$ translates the transform by $-h$.