The Student Room Group
Reply 1
hi, your question doesnt make sense, how can you prove |z+1+i/2|=rt(5)/2, when z is not given, and where does w fit in.
If you send me a valid question, i will be happy to answer it.
thanks
Reply 2
It's not wrong. The way I did it is in white.


w = u + vi = (z+2)/(z+i) = [(x+2)+iy]/[x+i(y+1)] = {[(x+2)+iy][x+i(y+1)]}/[x² + (y+1)²)] = [x(x+2) - (x+2)(y+1)i + xyi + y(y+1)]/[x² + (y+1)²)]

Since w lies on the imaginary axis, u=0
u = x(x+2) + y(y+1) = 0
=> + 2x + + y = 0
=> (x+1)² - 1 + (y+0.5)² - 0.25 = 0
=> (x+1)² + (y+0.5)² = (5/4)
So it's a circle with center (-1, -0.5) and radius 0.5sqrt[5], i.e.:
|z + 1 + 0.5i| = 0.5sqrt[5]
as required.

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