`z=-1+i=r(costheta + isintheta) rcostheta = -1, rsintheta = 1`

if |(2iz_1-z_1-z_2)/(2iz_1-z_1-z_2)|=|(costheta+ isintheta)/(costheta-isintheta)| then prove that z_1/z_2 is purely real.

Let z_1=r_1(costheta_1+isintheta_1)a n dz_2=r_2(costheta_2+isintheta_2) be two complex numbers. Then prove that |z_1+z_2|^2=r1 2+r2 2+2r_1r_2cos(theta_1-theta_2) or |z_1+z_2|^2=|z_1|^2+|z_2|^2+2|z_1||z_2|^()_cos(theta_1-theta_2) |z_1-z_2|^2=r1 2+r2 2-2r_1r_2cos(theta_1-theta_2) or |z_1-z_2|^2=|z_1|^2+|z_2|^2-2|z_1||z_2|^()_cos(theta_1-theta_2)

Show that : Costheta /( 1 - Sintheta ) + Costheta /( 1 + Sintheta ) = 2/Costheta

Evaluate : (( 1 + sintheta ) ( 1 - sintheta )) / ( ( costheta ) ( - costheta ) )

Prove that (frac(1+costheta+isintheta)(1+costheta-isintheta))^n=cos ntheta+isin ntheta .

If I= [[1,0],[0,1]] , J = [[0,1],[-1,0]] and B = [[costheta, sintheta],[-sintheta, costheta]] , then B= (A) Icostheta+Jsintheta (B) Icostheta-Jsintheta (C) Isintheta+Jcostheta (D) -Icostheta+Jsintheta

State true or false: If z= (cos2theta+isin2theta)/(costheta+isintheta) then z is unimodular.

If sqrt(3)-i=r(costheta+isintheta) , then r= . . . .