`|z_1+z_2|=|z_1-z_2| Leftrightarrow arg(z_1)-arg(z_2)=pi/2Leftrightarrow z_1/z_2` is purely imaginary

Prove that |z_1+z_2|^2=|z_1|^2, ifz_1//z_2 is purely imaginary.

If |z_1|=|z_2| and arg((z_1)/(z_2))=pi , then z_1+z_2 is equal to (a) 0 (b) purely imaginary (c) purely real (d) none of these

Given that |z_1+z_2|^2=|z_1|^2+|z_2|^2 , prove that z_1/z_2 is purely imaginary.

Prove that |z_1+z_2|^2=|z_1|^2+|z_2|^2, ifz_1//z_2 is purely imaginary.

If z_1, z_2 are two complex numbers (z_1!=z_2) satisfying |z_1^2-z_2^2|=| z_ 1^2+ z _2 ^2-2( z _1)( z _2)| , then a. (z_1)/(z_2) is purely imaginary b. (z_1)/(z_2) is purely real c. |a r g z_1-a rgz_2|=pi d. |a r g z_1-a rgz_2|=pi/2

If |z_(1)|=|z_(2)| and arg (z_(1))+"arg"(z_(2))=0 , then

if |z_1+z_2|=|z_1|+|z_2|, then prove that a r g(z_1)=a r g(z_2) if |z_1-z_2|=|z_1|+|z_2|, then prove that a r g(z_1)=a r g(z_2)=pi

If |z_(1)|=|z_(2)| and arg (z_(1)//z_(2))=pi, then find the of z_(1)z_(2).

Column I, Column II (Locus) parallelogram, p. z_1-z_4=z_2-z_3 rectangle, q. |z_1-z_(3|)=|z_2-z_(4|) rhombus, r. (z_1-z_2)/(z_3-z_4) is purely real square, s. (z_1-z_3)/(z_2-z_4) is purely imaginary , t. (z_1-z_2)/(z_3-z_2) is purely imaginary

Z_1!=Z_2 are two points in an Argand plane. If a|Z_1|=b|Z_2|, then prove that (a Z_1-b Z_2)/(a Z_1+b Z_2) is purely imaginary.