if `(1+i)z=(1-i)barz` then `z` is

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

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

If (1+i)z=(1-i) barz , then show that z=-i barz dot

IF (1+i)z=(1-i)z, then show that z=-barz .

Statement - I : If z = barz then z is purely imaginary Statement- II : If z = -barz then z is purely real

Conjugate of a complex no and its properties. If z, z_1, z_2 are complex no.; then :- (i) bar(barz)=z (ii)z+barz=2Re(z)(iii)z-barz=2i Im(z) (iv)z=barz hArr z is purely real (v) z+barz=0implies z is purely imaginary (vi)zbarz=[Re(z)]^2+[Im(z)]^2

If |z _1 |= 2 and (1 - i)z_2 + (1+i)barz_2 = 8sqrt2 , then the minimum value of |z_1 - z_2| is ______.

Let the lines (2-i)z=(2+i) barz and (2+i)z+(i-2)barz-4i=0 , (here i^(2)= -1 ) be normal to a circle C. If the line iz+barz+1+i=0 is tangent to this circle C, then its radius is :

If z= frac{4}{1-i} , then barz is (where barz is complex congugate of z)