The formula is correct, but it's normally given as:
period = 0.693×(R1 + 2R2)×C
Since frequency = 1/period and 1.44 = 1/0.693, your formula is equivalent to this.
R1 is the variable resistor in the schematic, R2 is the fixed resistor, and the capacitor C is the one attached to R2. Units are in ohms and farads.
ADDITIONAL NOTES:
In case you're wondering where the 0.693 comes from, it's the natural log of 2, and this derives from how the 555 timer operates, driving the capacitor voltage between 1/3 Vcc and 2/3 Vcc.
Be aware that in the example circuit, you cannot allow the adjustable resistor to go all the way to zero ohms, as this would then tie the DIS (discharge) pin of the 555 directly to the power supply. The first time the chip tries to discharge the capacitor, it will short out the power supply.
Using complex numbers to express sinusoidal signals is hardly "just a notational convenience".
On what it means for a sinusoid to have two orthogonal components:
First, realise that "orthogonal" is just a fancy word for "separate" or "fully independent".
Assume that you're dealing with a sinusoidal signal of fixed frequency \$ \omega \$. Such signals have two degrees of freedom - amplitude \$A\$ and phase \$\phi\$. That is:
$$
x(t) = \operatorname{Re}(A e^{j \phi}\times e^{j\omega t}) = A\cos ( \omega t + \phi )
$$
Information can be conveyed by either varying the amplitude or varying the phase, so there are two separate "channels" for information.
Equivalently, you can express the same fixed-frequency sinusoidal signal as the sum of two signals, phase-displaced by 90 degrees:
$$
x(t) = A_1 \sin (\omega t) + A_2 \cos (\omega t)
$$
Think of the sin term as a "vertical" wiggle, and the cos term as a "horizontal" wiggle. Again, these form two separate "channels" for communicating information.
It is fairly easy to build equipment that separates the sine component from the cosine component, so this is used as the basis of practical communications schemes. See quadrature amplitude modulation (QAM).
On the physical meaning of "multiplying by \$j\$":
In phasor form, the phase of the signal is given by a complex number \$e^{j\phi}\$ like so:
$$
e^{j\phi} = \cos{\phi} + j \sin \phi
$$
If you multiply by \$j\$ you get:
$$
j\times e^{j\phi}=j \cos \phi - sin \phi
$$
$$
j \times e^{j\phi} = j\sin (\phi+90^\circ)+cos(\phi + 90^\circ)
$$
$$
j \times e^{j\phi} = e^{j\phi+90^\circ}
$$
Which is to say that multiplying a phasor by \$j\$ changes its phase by \$+90^\circ\$. I like to think that two phasors \$A\$ and \$jA\$ are at right angles to each other, i.e. they are orthogonal.
Best Answer
Notice, that \$\forall x\in\mathbb{R}\$:
$$\left|\exp\left(xi\right)\right|=1\tag1$$
Because:
$$\left|\exp\left(xi\right)\right|=\left|\cos\left(x\right)+\sin\left(x\right)i\right|=\sqrt{\underbrace{\cos^2\left(x\right)+\sin^2\left(x\right)}_{=\space 1}}=\sqrt{1}=1\tag2$$
So, for your integral we get:
$$\int_0^{\text{T}_0}\left|\exp\left(\omega_0t\text{j}\right)\right|^2\space\text{d}t=\int_0^{\text{T}_0}\left|\cos\left(\omega_0t\right)+\sin\left(\omega_0t\right)\text{j}\right|^2\space\text{d}t=$$ $$\int_0^{\text{T}_0}\left(\sqrt{\underbrace{\cos^2\left(\omega_0t\right)+\sin^2\left(\omega_0t\right)}_{=\space1}}\right)^2\space\text{d}t=\int_0^{\text{T}_0}\underbrace{\left(\sqrt{1}\right)^2}_{=\space1}\space\text{d}t=$$ $$\int_0^{\text{T}_0}1\space\text{d}t=\left[t\right]_0^{\text{T}_0}=\text{T}_0-0=\text{T}_0\tag3$$