If -pi

Question

(A) Let $\arg (z) = 	heta$, where $-\pi < 	heta < -\frac{\pi}{2}$.Since $\bar{z}$ is the reflection of $z$ across the real axis, $\arg (\bar{z}) = -

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$\pi$

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(A) Let $\arg (z) = 	heta$, where $-\pi Since $\bar{z}$ is the reflection of $z$ across the real axis, $\arg (\bar{z}) = -\arg (z) = -	heta$.Since $-\pi Now, $-\bar{z} = -1 \cdot \bar{z} = e^{i\pi} \cdot \bar{z}$.Therefore, $\arg (-\bar{z}) = \arg (e^{i\pi}) + \arg (\bar{z}) = \pi + (-	heta) = \pi - 	heta$.We need to calculate $\arg (\bar{z}) - \arg (-\bar{z}) = -	heta - (\pi - 	heta) = -	heta - \pi + 	heta = -\pi$.Since the argument is typically defined in the range $(-\pi, \pi]$, we can add $2\pi$ to get the principal value: $-\pi + 2\pi = \pi$.

If $-\pi < \arg (z) < -\frac{\pi}{2}$, then $\arg (\bar{z}) - \arg (-\bar{z})$ is

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