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When calculating the real and imaginary parts of the complex number, do we take the angle as shown or the magnitude of it? I thought that we would just take the angle as shown, but apparently not, according to my textbook (unless its a typo):

$$z=2\sqrt{3}\operatorname{cis}\left(\frac{\pi }{3}\right)\:w=4\operatorname{cis}\left(\frac{-\pi }{6}\right)$$

Find $z + w$ in:

  1. Cartesian form
  2. Modulus-argument form

I worked out the complex number to be $3\sqrt{3}+i$:

$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:=\:2\sqrt{3}-2i\\ 2\sqrt{3}\cos\left(\frac{\pi }{3}\right)+2\sqrt{3}i\:\sin\left(\frac{\pi }{3}\right)=\:\sqrt{3}\:+3i\\ \sqrt{3}\:+3i\:+\:2\sqrt{3}-2i\:=\:3\sqrt{3}+i$$

(Following from this I got the mod-arg form to be $[2\sqrt{7},0.19]$)

The textbook says pretty much the same thing, except $$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:$$ is taken to be: $4\cos\left(\frac{\pi }{6}\right)+4i\sin\left(\frac{\pi }{6}\right)$, resulting in the complex number $=3\sqrt{3}+5i$.

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  • $\begingroup$ What's $cis$? Is it a typo of $\cos$? $\endgroup$ – 5xum Aug 29 '18 at 14:34
  • $\begingroup$ $cis (x) = cos (x) + i sin(x)$ $\endgroup$ – Chx Aug 29 '18 at 14:35
  • $\begingroup$ @CheksNweze For mathjax, try $\operatorname{cis}$ and $\cos$, $\sin$ $\endgroup$ – Kevin Aug 29 '18 at 14:36
  • $\begingroup$ @Kevin sorry, what do you mean? $\endgroup$ – Chx Aug 29 '18 at 14:37
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    $\begingroup$ @CheksNweze I had hoped that the syntax for mathjax would show, but, your operators in the OP are in standard font, not mathematical typeface. To get them in the latter, type (enclosed in dollar signs) "\operatorname{cis}", "\sin" and "\cos". $\endgroup$ – Kevin Aug 29 '18 at 14:45
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We take the angle as shown, not just its absolute value. Therefore, $4\operatorname{cis}\left(\frac\pi6\right)\neq4\operatorname{cis}\left(-\frac\pi6\right)$. It looks as if you are right and your textbook is wrong.

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You are correct and the text book is wrong.

$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:=\:2\sqrt{3}-2i\\ 2\sqrt{3}\cos\left(\frac{\pi }{3}\right)+2\sqrt{3}i\:\sin\left(\frac{\pi }{3}\right)=\:\sqrt{3}\:+3i\\ \sqrt{3}\:+3i\:+\:2\sqrt{3}-2i\:=\:3\sqrt{3}+i$$ is correct.

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