MHB Sum of two inverse tangent functions

[DreamWeaver]

By considering the product of complex numbers:

$$z = (2+i)(3+i)$$

Show that

$$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \frac{\pi}{4}$$

 

[polygamma]

Spoiler

$$(2+i)(3+i) = 5+5i = \sqrt{50}e^{i \arctan (1)} = \sqrt{50}e^{i \frac{\pi}{4}}$$

$$ (2+i)(3+i) = \sqrt{5} e^{i \arctan (\frac{1}{2})} \sqrt{10} e^{i \arctan (\frac{1}{3})} = \sqrt{50} e^{i [\arctan (\frac{1}{2}) + i \arctan(\frac{1}{3})]}$$

Therefore,

$$ \frac{\pi}{4} = \arctan \left( \frac{1}{2}\right) + \arctan \left(\frac{1}{3} \right)$$

 

[DreamWeaver]

Spoiler

$$(2+i)(3+i) = 5+5i = \sqrt{50}e^{i \arctan (1)} = \sqrt{50}e^{i \frac{\pi}{4}}$$

$$ (2+i)(3+i) = \sqrt{5} e^{i \arctan (\frac{1}{2})} \sqrt{10} e^{i \arctan (\frac{1}{3})} = \sqrt{50} e^{i [\arctan (\frac{1}{2}) + i \arctan(\frac{1}{3})]}$$

Therefore,

$$ \frac{\pi}{4} = \arctan \left( \frac{1}{2}\right) + \arctan \left(\frac{1}{3} \right)$$

Very efficient! But then, I'd expect nothing less from you. (Yes)