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If |z1-z0|=z2-z1=pi/2 , then find z0....

If `|z_1-z_0|=z_2-z_1=pi/2` , then find `z_0`.

Text Solution

Verified by Experts

The correct Answer is:
`(1)/(2){(i +1)z_(1) +(1-i)z_(2)}`
Here, `|(z_(2)-z_(0))/(z_(0)-z_(1))| = 1 and amp ((z_(2)-z_(0))/(z_(0)-z_(1))) = (pi)/(2)`
`therefore (z_(2)-z_(0))/(z_(0)-z_(1)) = 1{cos.(pi)/(2) + i sin .(pi)/(2)} =i`
`rArr z_(2) -z_(0) = iz_(0) = iz_(0)`
`rArr z_(2)+iz_(1) = (i+1)z_(0)`
`or z_(0) = (z_(2) + iz_(1))/(1+i)`
`=((z_(2) + iz_(1))(1-i))/(1^(2)+ 1^(2))`
`=(1)/(2){(i+ 1)z_(1)+(1-i)z_(2)}`
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