How can the polar form of a complex number be$r(\cos(x)-i\sin(x))$?

I don't understand how this can form can work: $r(\cos(x)-i\sin(x))$. I saw it in my textbook. Surely there should be a "$+$" rather than a "$-$" between the $\cos$ and the $\sin$. If the imaginary part was negative then surely that would be accounted for by the argument which in this case I have written as "$x$".

• you are right, it should be $+$ Sep 14 '14 at 15:07
• Perhaps they were writing $\bar z$, the complex conjugate of $z$? Sep 14 '14 at 15:09
• You can write a complex number either as $r(\cos x - i \sin x)$ or as $r(\cos\theta + i \sin\theta)$, simply by substituting $x=-\theta$. Sep 14 '14 at 15:09
• Where in what book? Sep 14 '14 at 15:22
• IF you gave an exact quote from the textbook, including context, we might be able to help. Sep 14 '14 at 15:23

Perhaps you could use the fact that $-\sin (x) = \sin (-x)$ and $\cos (x) = \cos (-x)$, then your expression becomes $r(\cos (-x) + i \sin (-x))$.