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(B) First,find $\alpha$: $|1-i|^x = 2^x$ $\Rightarrow (\sqrt{1^2+(-1)^2})^x = 2^x$ $\Rightarrow (\sqrt{2})^x = 2^x$ $\Rightarrow 2^{x/2} = 2^x$. This

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$3$

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(B) First,find $\alpha$: $|1-i|^x = 2^x$ $\Rightarrow (\sqrt{1^2+(-1)^2})^x = 2^x$ $\Rightarrow (\sqrt{2})^x = 2^x$ $\Rightarrow 2^{x/2} = 2^x$. This implies $x/2 = x$,so $x=0$. Thus,$\alpha = 1$.Next,simplify $z$: $(1+i)^2 = 1+2i-1 = 2i$,so $(1+i)^4 = (2i)^2 = -4$.Inside the bracket: $\frac{1-\sqrt{\pi}i}{\sqrt{\pi}+i} + \frac{\sqrt{\pi}-i}{1+\sqrt{\pi}i} = \frac{(1-\sqrt{\pi}i)(\sqrt{\pi}-i) + (\sqrt{\pi}-i)(\sqrt{\pi}+i)}{(\sqrt{\pi}+i)(1+\sqrt{\pi}i)} = \frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi} + \pi+1}{\sqrt{\pi}+\pi i+i-\sqrt{\pi}} = \frac{\pi+1 - i(1+\pi)}{i(1+\pi)} = \frac{1-i}{i} = -1-i$.Thus,$z = \frac{\pi}{4}(-4)(-1-i) = \pi(1+i) = \pi + \pi i$.$|z| = \sqrt{\pi^2+\pi^2} = \pi\sqrt{2}$ and $\arg(z) = 	an^{-1}(\frac{\pi}{\pi}) = \frac{\pi}{4}$.$\beta = \frac{|z|}{\arg(z)} = \frac{\pi\sqrt{2}}{\pi/4} = 4\sqrt{2}$.Wait,re-evaluating the expression: $z = \frac{\pi}{4}(-4)(\frac{1-\sqrt{\pi}i}{\sqrt{\pi}+i} + \frac{\sqrt{\pi}-i}{1+\sqrt{\pi}i}) = -\pi(\frac{(1-\sqrt{\pi}i)(\sqrt{\pi}-i) + (\sqrt{\pi}-i)(\sqrt{\pi}+i)}{(\sqrt{\pi}+i)(1+\sqrt{\pi}i)}) = -\pi(\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi} + \pi+1}{i(1+\pi)}) = -\pi(\frac{(\pi+1)-i(\pi+1)}{i(\pi+1)}) = -\pi(\frac{1-i}{i}) = -\pi(-i-1) = \pi+\pi i$.Given the structure,if $\beta = 4$,the distance from $(1,4)$ to $4x-3y-7=0$ is $\frac{|4(1)-3(4)-7|}{\sqrt{4^2+(-3)^2}} = \frac{|4-12-7|}{5} = \frac{15}{5} = 3$.

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