Judging by the given Argand diagram, it looks like
[tex]v = 2e^{i5\pi/12}[/tex]
[tex]w = 4e^{i9\pi/12} = 4e^{i3\pi/4}[/tex]
where I assume the innermost circle has radius 1, and the radius of each subsequent circle increases by 1. There are 12 equally spaced rays spanning the positive imaginary half-plane, so there's an angle of [tex]\frac\pi{12}[/tex] between each adjacent pair of rays.
Subtraction is best carried out in rectangular form.
Since [tex]v[/tex] is in the first quadrant, we know both [tex]\cos\left(\frac{5\pi}{12}\right)[/tex] and [tex]\sin\left(\frac{5\pi}{12}\right)[/tex] are positive. Then
[tex]\cos\left(\dfrac{5\pi}{12}\right) = \sqrt{\dfrac{1 + \cos\left(\frac{5\pi}6\right)}2} = \dfrac{\sqrt{2 - \sqrt3}}2[/tex]
and
[tex]\sin\left(\dfrac{5\pi}{12}\right) = \sqrt{\dfrac{1 - \cos\left(\frac{5\pi}6\right)}2} = \dfrac{\sqrt{2 + \sqrt3}}2[/tex]
Suppose we can denest the roots and write
[tex]\sqrt{2 \pm \sqrt3} = a + b \sqrt c[/tex]
Taking squares on both sides gives
[tex]2 \pm \sqrt3 = a^2 + b^2c + 2ab \sqrt c[/tex]
Let [tex]c=3[/tex] and [tex]2ab=\pm1[/tex], so [tex]b=\pm\frac1{2a}[/tex]. Then
[tex]a^2+b^2c = a^2 + \dfrac 3{4a^2} = 2 \implies 4a^4 - 8a^2 + 3 = 0 \\\\ \implies (2a^2 - 3) (2a^2 - 1) = 0[/tex]
Solving for [tex]a[/tex] and [tex]b[/tex], we get
[tex]a = \pm \sqrt{\dfrac32} \implies b = \pm\dfrac1{\sqrt6}[/tex]
or
[tex]a = \pm \dfrac1{\sqrt2} \implies b = \pm\dfrac1{\sqrt2}[/tex]
Either way, we end up with
[tex]\sqrt{2 \pm \sqrt3} = \sqrt{\dfrac32} + \dfrac{\sqrt3}{\sqrt6} = \dfrac{\sqrt3 \pm 1}{\sqrt2}[/tex]
Now,
[tex]v = 2\left(\cos\left(\dfrac{5\pi}{12}\right) + i \sin\left(\dfrac{5\pi}{12}\right)\right) \\\\ ~~~~= \sqrt{2 - \sqrt3} + i \sqrt{2 + \sqrt3} \\\\ ~~~~ = \dfrac{\sqrt3 - 1}{\sqrt2} + i \dfrac{\sqrt3 + 1}{\sqrt2}[/tex]
and
[tex]w = 4\left(\cos\left(\dfrac{3\pi}4\right) + i \sin\left(\dfrac{3\pi}4\right)\right) \\\\ ~~~~ = -2\sqrt2 + i\,2\sqrt2[/tex]
so that
[tex]v - w = \dfrac{\sqrt3 + 3}{\sqrt2} + i \dfrac{\sqrt3-3}{\sqrt2}[/tex]
Compute the modulus.
[tex]r_1 = |v - w| \\\\ ~~~~ = \sqrt{\left(\dfrac{\sqrt3 + 3}{\sqrt2}\right)^2 + \left(\dfrac{\sqrt3 - 3}{\sqrt2}\right)^2} \\\\ ~~~~ = \sqrt{\dfrac{24}2} = \sqrt{12} = \boxed{2\sqrt3}[/tex]
Compute the argument. Observe that [tex]v+w[/tex] would end up in the second quadrant, so [tex]v-w[/tex] lies opposite this point in the fourth quadrant.
[tex]\theta_1 = \tan^{-1}\left(\dfrac{\frac{\sqrt3-3}{\sqrt2}}{\frac{\sqrt3+3}{\sqrt2}}\right) \\\\ ~~~~ = \tan^{-1}\left(\sqrt3 - 2\right) = \boxed{-\dfrac\pi{12}}[/tex]
which follows from the observation that
[tex]\tan\left(\dfrac{5\pi}{12}\right) = \dfrac{\sin\left(\frac{5\pi}{12}\right)}{\cos\left(\frac{5\pi}{12}\right)} = \dfrac{\sqrt{2 + \sqrt3}}{\sqrt{2 - \sqrt3}} = 2 + \sqrt3[/tex]
and
[tex]\tan\left(\dfrac{5\pi}{12}\right) = \tan\left(-\dfrac\pi{12} + \dfrac\pi2\right) = -\cot\left(-\dfrac\pi{12}\right) = \dfrac1{\tan\left(\dfrac\pi{12}\right)}[/tex]
which together give
[tex]\tan\left(\dfrac\pi{12}\right) = \dfrac1{2 + \sqrt3} = 2 - \sqrt3 = -(\sqrt3 - 2) \\\\ \implies \tan\left(-\dfrac\pi{12}\right) = \sqrt3 - 2 \\\\ \implies -\dfrac\pi{12} = \tan^{-1}(\sqrt3-2)[/tex]
