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What is the argument of $z = (1+\cos 2a)+i(\sin 2a)$ if $\pi/2<a<3\pi/2 $?

After using the formula of $\sin2a$ and $\cos2a$, I am getting the argument as $a$ when $\pi/2<a<\pi$ and $a-2\pi$ when $\pi<a<3\pi/2$ but both the answers are incorrect

My approach my approach

Answer given in book answer in book

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  • $\begingroup$ Could you type in your approach. $\endgroup$ – Mohammad Zuhair Khan Jan 18 '18 at 17:38
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    $\begingroup$ Why you think the book's answer is incorrect? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 18 '18 at 17:51
  • $\begingroup$ It is important to realize that $tana=\frac{sin2\alpha}{1+cos2\alpha}$, however, arctan returns angles between $-\pi/2$ and $\pi/2$, but your given domain for $a$ falls outside the arctan range. See for yourself with a numerical example, say $a=2\pi/3$. Arctan returns $-\pi/3$. And that is rectified in the last two lines of the book's answer $\endgroup$ – imranfat Jan 18 '18 at 17:55
  • $\begingroup$ See en.wikipedia.org/wiki/Atan2#Definition_and_computation $\endgroup$ – lab bhattacharjee Jan 19 '18 at 5:05
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I am giving a different approach here, less formal but with the intention to see what is going on and why the argument cannot be $a$, as the OP understandably suggested. Without a bit "digging" into the numbers, it is understandable that an average reader (like me!) wouldn't immediately follow the last to lines of the book's answer as to why these steps are needed.

When we let $a$ run from interval $\pi/2$ to $3\pi/2$ and we calculate the arguments (I made corresponding "lists" with the TI), the corresponding interval for arg(z) is from $-\pi/2$ to $\pi/2$. When you want to relate this with input $a$, it becomes clear that $a$ is $\pi$ "too high", from which the book's suggestion $arg(z)=a-\pi$ follows.

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HINT

Note that

$$\frac{\sin 2a}{1+\cos 2a}=\tan a$$

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By your work it's $$2\cos\alpha(\cos\alpha+i\sin\alpha)=-2\cos\alpha(\cos(\alpha-\pi)+i\sin(\alpha-\pi)).$$ I think by the definition in the book $$-\pi<\arg{z}\leq\pi$$ and since $$-\frac{\pi}{2}<\alpha-\pi<\frac{\pi}{2},$$ we obtain $\arg{z}=\alpha-\pi.$

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  • $\begingroup$ You are correct, but this may not be immediately evident by the casual reader, hence my less rigorous explanation! $\endgroup$ – imranfat Jan 18 '18 at 18:23

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