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Mathematics Question on complex numbers

Let S be the set of all complex numbers z satisfying ∣z−2+i∣≥5.\left|z-2+i\right|\ge\sqrt{5}. If the complex number z0z_0 is such that 1∣z0−1∣\frac{ 1}{\left|z_{0}-1\right|} is the maximum of the set \left\\{\frac{ 1}{\left|z_{0}-1\right|}: z\,\in\,S\right\\}, then the principal argument of 4−z0−z0−z0−z0−+2i\frac{4-z_{0}-z^{-}_{0}}{z_{0}-z^{-}_{0}+2i} is

A

−π2-\frac{\pi}{2}

B

π4\frac{\pi}{4}

C

π2\frac{\pi}{2}

D

−3π4-\frac{3\pi}{4}

Answer

−π2-\frac{\pi}{2}

Explanation

Solution

∣z−2+i∣≥5\left|z-2+i\right|\ge\sqrt{5} ∣1z0−1∣\left|\frac{1}{z_{0}-1}\right| is maximum when ∣z0−1∣|z_{0} - 1| is minimum Let z0=x+iyz_{0}=x+iy x<1x <1 and y>0y>0 4−z0−z0ˉz0−z0+2iˉ\frac{4-z_{0}-\bar{z_{0}}}{z_{0}-\bar{z_{0}+2i}} =4−x−iy−x−iyx+iy−x+iy+2i=\frac{4-x-iy-x-iy}{x+iy-x+iy+2i} =4−2x(y+1)2i=−i(2−x)(y+1)=\frac{4-2x}{\left(y+1\right)2i}=\frac{-i\left(2-x\right)}{\left(y+1\right)} ∵2−xy+1\because \frac{2-x}{y+1} is a positive real number ⇒arg(4−z0−z0z0−z0+2i)=−π2\Rightarrow arg \left(\frac{4-z_{0}-z_{0}}{z_{0}-z_{0}+2i}\right)=-\frac{\pi}{2}