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Problem: For a number $z\in\mathbb{C},$ the following conditions apply:

  1. $z+\bar{z}>0.$

  2. $iz+\bar{iz}<0.$

Determine in which quadrant $z$ lies.

Attempt: I simply substituted $z=a+bi, \quad a,b>0$ and computed both conditions.

  1. $$a+bi+a+bi=a+a+bi-bi=2a>0.$$
  2. $$i(a+bi)+\overline{i(a+bi)}=-b+ai+\overline{(-b+ai)}=-b+ai-b-ai=-2b<0.$$

Since these 2 conditions are satisfied when the real numbers $a,b>0$, it means that $z$ has to exist in the first quadrant. Is my reasoning correct? Any other method or improvement?

Parseval
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1 Answers1

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The answer is quite simple when you realize that

$$\arg Z=\frac{1}{2i}\ln\frac{Z}{Z^*}$$

Thus, for the first case, with $Z=z+z^*$ we obtain

$$\arg Z=\frac{1}{2i}\ln\frac{z+z^*}{z^*+z}=\frac{1}{2i}\ln(1)=0$$

and the result lies on the positive $x$-axis (so I guess you'd say the $1^{st}$ quadrant).

Similarly, for the second case, with $Z=iz+iz^*$ we obtain

$$\arg Z=\frac{1}{2i}\ln\frac{iz+iz^*}{-iz^*-iz}=\frac{1}{2i}\ln(-1)=\frac{2\pi i}{2i}=\pi$$

and the result lies on the negative$x$-axis (so I guess you'd say the $2^{nd}$ or $3^{rd}$ quadrant).

Notice that I did not have to make any recourse to the Cartesian form $z=a+ib~$ or the polar form $z=re^{i\theta}$.

Cye Waldman
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  • I have just determined that saying $\arg Z=\frac{1}{2i}\ln\frac{Z}{Z^}$ can get you into the same problems as taking $\tan$ as opposed to $\tan2$ in some situations. To be on the safe side, you should always use $\arg Z=\frac{1}{2i}(\ln{Z}-\ln{Z^})$. – Cye Waldman Aug 24 '17 at 21:03