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Let z1=r1(costheta1+isintheta1) and z2=r...

Let `z_1=r_1(costheta_1+isintheta_1)` and `z_2=r_2(costheta_2+isintheta_2)` be two complex numbers then prove the following

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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)

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)

|1+costheta+isintheta|=

(1-costheta+isintheta)^8 =

(1-costheta+isintheta)^6 =

(1+costheta-isintheta)^4 =

(1-costheta+isintheta)^(6)

(1+costheta-isintheta)^(n)

Convert z=2/(1+costheta+isintheta) in a+ib form

If (1+2i)/(2+i)=r(costheta+isintheta) then