If alpha is the modulus of z_1=4+3 i,then a p | vedclass

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(B) Given $z_1 = 4+3 i$. The modulus $\alpha = |z_1| = \sqrt{4^2+3^2} = \sqrt{16+9} = 5$. The region is defined by $|z - \overline{z_1}| \leq 5$. Sinc

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$z_1$

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(B) Given $z_1 = 4+3 i$. The modulus $\alpha = |z_1| = \sqrt{4^2+3^2} = \sqrt{16+9} = 5$. The region is defined by $|z - \overline{z_1}| \leq 5$. Since $\overline{z_1} = 4-3 i$,the inequality is $|z - (4-3 i)| \leq 5$. We test the options by checking if they satisfy $|z - (4-3 i)| \leq 5$: For option $B$,$z = z_1 = 4+3 i$: $|(4+3 i) - (4-3 i)| = |6 i| = 6$. Since $6 > 5$,the point $z_1$ does not satisfy the inequality. Thus,$z_1$ does not lie in the region.

If $\alpha$ is the modulus of $z_1=4+3 i$,then a point that does not lie in the region represented by $|z-\overline{z_1}| \leq \alpha$ is

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