`z_1`=2-i, `z_2`=1+i, find `abs(frac{z_1+z_2+1}{z_1-z_2+1})`

Let z_1 =2-i, z_2 = -2 +i, find Re[ frac{z_1 z_2}{z_1} ]

Let z_1 =2-i, z_2 = -2 +i, find Im[ frac{1}{z_1 z_2} ]

If z_1 = 1-i and z_2 = 2+4i , then im[frac{z_1z_2}{z_1} =

If abs(z_1) =1, ( z_1 != -1) and z_2 = frac{z_1 - 1}{z_1 +1} , then show that the real part of z_2 is zero.

If z_1, z_2, z_3 are complex numbers such that abs(z_1)= abs(z_2) = abs(z_3) = abs(frac{1}{z_1}+frac{1}{z_2}+frac{1}{z_3}) = 1 , then abs(z_1+z_2+z_3) is

If frac{2z_1}{3z_2} is a purely imaginary number, then abs(frac{z_1-z_2}{z_1+z_2}) =

If z_1=1-2i , z_2=1+i and z_3=3+4i , then [frac{1}{z_1}+frac{3}{z_2}][frac{z_3}{z_2}=

Let z1 be a complex number with abs(z_1)=1 and z2 be any complex number, then abs(frac{z_1-z_2}{1-z_1barz_2}) =

Let z_1 =3+4i and z_2 = -1+2i . Then abs(z_1+z_2)^2-2(abs(z_1)^2+abs(z_2)^2) is equal to

If z=x+iy then abs(frac{z-1}{z+1}) =1 represents