# Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$

Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$

$|(1+i\sqrt{3})(1+i)|=8$

Using Moivre's theorem on the LHS:

$$\sqrt{(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)}=2\sqrt{2}\left(\cos\frac{\phi+2k\pi}{2}+i\sin\frac{\phi+2k\pi}{2}\right),k=0,1$$

This is not correct (checked for $\phi=\pi/3$).

How to prove this equation?

• The statement $(1+i\sqrt{3})(1+i)=8$ is clearly false. – copper.hat Oct 27 '15 at 15:58
• The statement is still false, you are missing a square. – copper.hat Oct 27 '15 at 16:00
• @copper.hat Is the Moivre's theorem the right approach for this proof? – user300045 Oct 27 '15 at 16:02
• I have no idea what you are doing with the $k$ above. Look at Math's answer below. The key to that answer is the fact that ${7 \pi \over 12} = {\pi \over 3} + {\pi \over 4}$. – copper.hat Oct 27 '15 at 16:08
• What is $\sqrt{z}$ when, like here, $z$ is a complex number which is not positive real? – Did Oct 27 '15 at 16:08

We have $$(1+i\sqrt{3})=2(\cos \frac{\pi}{3}+i\sin\frac{\pi}{3}),$$ $$(1+i)=\sqrt{2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4}).$$ Then the product of these is $2\sqrt{2}(\cos \frac{7\pi}{12}+i\sin\frac{7\pi}{12})$. So $$2\sqrt{2}(\cos \frac{7\pi}{12}+i\sin\frac{7\pi}{12})(\cos {\phi}+i\sin{\phi})=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right).$$
Since you can write $(1+i\sqrt{3})(1+i)e^{i \phi}=2\sqrt{2}e^{i \phi} e^{i {7 \pi \over 12}}$, so you can concentrate on showing $(1+i\sqrt{3})(1+i)=2\sqrt{2} e^{i {7 \pi \over 12}}$ instead.