The inverse point of 1+i with respect to the circle |z-1|=2 is
Formula for symmetric point with respect to the circle-
∣z−zo∣=r|z-z_o|=r∣z−zo∣=r
So, z′=r2zˉ−zo+zo −(1)z'=\dfrac{r^2}{\bar{z}-z_o}+z_o~~~~~~~~~-(1)z′=zˉ−zor2+zo −(1)
Given circle is-
∣z−1∣=2|z-1|=2∣z−1∣=2
So, zo=1,r=2 and point z=1+i\text{ So, } z_o=1,r=2 \text{ and point }z=1+i So, zo=1,r=2 and point z=1+i
Subsititue above values in (1)-
z′=221+i‾−1=41−i−1=4−i=4iz'=\dfrac{2^2}{\overline{1+i}-1}=\dfrac{4}{1-i-1}=\dfrac{4}{-i}=4iz′=1+i−122=1−i−14=−i4=4i
Hence The inverse is 4i.
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