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In complex numbers , when the argument of a complex number lies in the second quadrant we usually say that it’s argument is $\pi -\theta$ however what is the meaning of the notation $\theta-\pi$ ? For example , if I wanted to write the modulus argument form of $1+ i\tan\theta$ when $\theta$ lies in the second quadrant I would usually write it as $-\sec\theta \space\text{cis}\space(\pi-\theta)$ however the notation of $-\sec\theta\space\text{cis}(\theta-\pi)$ is also valid . How are both the same ?

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$1+i\tan \theta =\sec \theta(\cos \theta + i\sin \theta)=\sec \theta \space \text{cis}\space \theta\\\hspace{43pt}=\sec \theta (-\cos(\pi-\theta)+i\sin(\pi-\theta))\\\hspace{43pt}= -\sec \theta (\cos (\pi-\theta) -i\sin(\pi-\theta))\\\hspace{43pt} =-\sec \theta(\cos(\theta-\pi)+i\sin(\theta-\pi))\\\hspace{43pt} =-\sec\theta \space \text{cis}\space(\theta-\pi)$


$-\sec \theta \space\text{cis}\space (\pi - \theta) = -\sec \theta(\cos(\pi-\theta)+i\sin(\pi-\theta))\\\hspace{75pt}= -\sec \theta (-\cos \theta+i\sin \theta)\\\hspace{75pt}=\sec\theta(\cos \theta-i\sin\theta)\\\hspace{75pt}=\sec \theta\space\text{cis}\space(-\theta) \ne \sec \theta \space \text{cis}\space \theta\\\hspace{75pt}=1-i\tan\theta \ne 1+i\tan \theta $

So, they're not actually the same.


As $\theta\in (\frac\pi2,\pi)$,

($\pi-\theta$) means $Q_1$ so imaginary part of $-\sec\theta \space \text{cis}\space(\theta-\pi)$ would lie in $Q_4$.

($\theta-\pi$) means $Q_4$ so imaginary part of $-\sec\theta \space \text{cis}\space(\pi-\theta)$ would lie in $Q_1$.

Also, imaginary part of $1+i\tan \theta$, which is $\tan \theta$, lies in $Q_1$ so that verifies this.

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  • $\begingroup$ desmos.com/calculator/n1698xdz63 $\endgroup$ – Maadhav Gupta Jan 16 '18 at 6:51
  • $\begingroup$ Thank you , but I’m the second step, why did you write $-cos(\pi-\theta)$ if we know that $cos(\pi-\theta)$ is already negative ? $\endgroup$ – Aditi Jan 16 '18 at 6:58
  • $\begingroup$ $\cos \theta$ lies in $Q_2$ so it would be negative but $\cos (\pi - \theta)$ lies in $Q_1$ so as a result, it would become positive. So we need the minus to maintain the negative sign. $\endgroup$ – Maadhav Gupta Jan 16 '18 at 7:05
  • $\begingroup$ I’m sorry but I don’t understand the meaning. If $\theta$ lies in second quadrant then how does $\pi -\theta$ lie in first quadrant ? I’m not able to imagine it actually $\endgroup$ – Aditi Jan 16 '18 at 7:11
  • $\begingroup$ $\theta \in (90^\circ,180^\circ)$. So, ($\pi - \theta) = (180^\circ-\theta)\in(0^\circ, 90^\circ)$. $\endgroup$ – Maadhav Gupta Jan 16 '18 at 7:15

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