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


Using Moivre's theorem on the LHS:


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

How to prove this equation?

  • 1
    $\begingroup$ The statement $(1+i\sqrt{3})(1+i)=8$ is clearly false. $\endgroup$ – copper.hat Oct 27 '15 at 15:58
  • $\begingroup$ The statement is still false, you are missing a square. $\endgroup$ – copper.hat Oct 27 '15 at 16:00
  • $\begingroup$ @copper.hat Is the Moivre's theorem the right approach for this proof? $\endgroup$ – user300045 Oct 27 '15 at 16:02
  • $\begingroup$ 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 ${7 \pi \over 12} = {\pi \over 3} + {\pi \over 4}$. $\endgroup$ – copper.hat Oct 27 '15 at 16:08
  • $\begingroup$ What is $\sqrt{z}$ when, like here, $z$ is a complex number which is not positive real? $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.