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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Jan 2008 08:52:42 IST
Accepted Answer [?]
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Method 1,
w = 1 - iz / z - i
lwl = 1
l z - i l = l(1 - iz)l = lz +1l
So z lies on the right bisector of line joining points 0 + i and 0 - i
Hence z lies on real axis.
Method 2,
lwl = 1
l1 - izl/lz - il = 1
So l1 - izll^2 = lz - il^2
 (1 - iz)(1 + izbar) = (z - i)(z bar + i)
Let bar be represented by '
Now (iz)' = -iz'
1 - iz + (iz)' +zz' = zz' + iz - iz' + 1
2 i ( z - z') = 0
Hence 2i (2iy) = 0
and y = 0..........the equatn of real axis
Hope you got it.Please nudge if not clear.Rate if satisfied.
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