`If (1+i)z = (1-i)barz, "then show that "barz = -ibarz.`

If (1+i)z = (1-i)barz, "then show that "z = -ibarz.

If (1+i)z=(1-i)barz , then z is equal to

If z= x+iy, "then show that " z barz + 2(z + barz) +b = 0, where b ∈R,"respresents a circle" .

If z = x + iy , then show that 2 bar(z) + 2 (a + barz) + b = 0 , where b in R , represents a circles.

z_(1) "the"z_(2) "are two complex numbers such that" |z_(1)| = |z_(2)| . "and" arg (z_(1)) + arg (z_(2) = pi," then show that "z_(1) = - barz_(2).

Solve: z +2barz=ibarz

if z=3 -2i, then verify that (i) z + barz = 2Rez (ii) z - barz = 2ilm z

if z = 2 + i + 4i^(2) -6i^(3) then verify that (i) (bar(z^(2)) = (barz)^(2))

If z_(1) = 3+ 2i and z_(2) = 2-i then verify that (i) bar(z_(1) + z_(2)) = barz_(1) + barz_(2)

Let (2-i)z=(2+i)barz and (2+i)z +(-2+i)barz-i=0 be normal to the circle and iz+barz+1+i=0 is tangent to the same circle having radius r . Then vaue of 128r^2