Question -
Answer -
|z1| = |z2|and arg (z1) + arg (z2) = π
Let us assume arg (z1)= θ
arg (z2) =π – θ
We know that in thepolar form, z = |z| (cos θ + i sin θ)
z1 =|z1| (cos θ + i sin θ) …………. (i)
z2 =|z2| (cos (π – θ) + i sin (π – θ))
= |z2|(-cos θ + i sin θ)
= – |z2|(cos θ – i sin θ)
Now let us find theconjugate of
= – |z2|(cos θ + i sin θ) …… (ii) (since, |Z2¯¯¯¯¯¯|=|Z2|)
Now,
z1 /= [|z1| (cos θ + i sinθ)] / [-|z2| (cos θ + i sin θ)]
= – |z1| /|z2| [since, |z1| = |z2|]
= -1
When we cross multiplywe get,
z1 = –