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If the expression $(1 + ir)^3$ is of the form of $s(1 + i)$ for some real $s$ where $r$ is also real and then the value of $r$ can be

$(A) \cot{\frac{\pi}{8}}$ $(B) \tan{\frac{\pi}{12}}$ $(C) \tan{\frac{5\pi}{8}}$ $(D) \sec{\pi}$

Since real and imaginary parts of the given complex number should be same. So I put $3\tan^{-1}(r)=\pi/4$ I'm only getting (B) as an answer but according to my book (B),(C),(D) are all correct options. Where did I go wrong?

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  • $\begingroup$ There is no answer (D) ! $\endgroup$ – Yves Daoust Jul 26 '16 at 8:45
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The argument of $(1+ir)^3$ is $3$ times that of $1+ir$, which is known to be one of $\pi/4$, $5\pi/4$. Then $$\arctan(r)=\frac\pi{12},\frac{9\pi}{12},\frac{17\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{21\pi}{12}.$$

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  • $\begingroup$ Wait wait...I don't get you.The solutions should be $3\tan^{-1}(r)=\pi/4+2\pi n$ right? $\endgroup$ – user220382 Jul 26 '16 at 8:57
  • $\begingroup$ @ZOZ: $s$ can be negative. $\endgroup$ – Yves Daoust Jul 26 '16 at 9:02
  • $\begingroup$ Oh silly me.I get it now!Thanks! $\endgroup$ – user220382 Jul 26 '16 at 9:02
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HINT:

$$1-3r^2+i(3r-r^3)=s(1+i)$$

$$\implies3r-r^3=1-3r^2$$

If $r=\tan A\implies\tan3A=\dfrac{3r-r^3}{1-3r^2}=1\implies3A=m\pi+\dfrac\pi4=(4m+1)\dfrac\pi4$ where $m$ is any integer

$A=(4m+1)\dfrac\pi{12}$ where $m\equiv0,1,2\pmod3$

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  • $\begingroup$ Nice...but can you tell me the mistake in my method?Why is my method not giving all the solutions? $\endgroup$ – user220382 Jul 26 '16 at 8:26
  • $\begingroup$ @ZOZ, (Sanchayan) You are taking the principal root only. $\endgroup$ – lab bhattacharjee Jul 26 '16 at 8:27
  • $\begingroup$ So what should be the correct condition in place of $3\tan^{-1}|r|=\pi/4$ ? $\endgroup$ – user220382 Jul 26 '16 at 8:28
  • $\begingroup$ Eh!You revealed my pseudonym :-P $\endgroup$ – user220382 Jul 26 '16 at 8:30
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    $\begingroup$ @ZOZ, Yes, $3\tan^{-1}r=n\pi+\dfrac\pi4$ $\endgroup$ – lab bhattacharjee Jul 26 '16 at 8:32

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