Let z 1 = a 1 + ib 1 (a 1 , b 1 ) and z 2 = a 2 + ib 2 (a 2 , b 2 ); where i = -1 , be…

Let z₁ = a₁ + ib₁ \(\equiv\)(a₁, b₁) and z₂ = a₂ + ib₂ \(\equiv\) (a₂, b₂); where \(\mathrm{i}=\sqrt{-1},\) be two complex numbers.
 
If ∠POQ = θ, From Rotation theorem
\(\frac{z_{2}-0}{z_{1}-0}=\frac{\left|z_{2}\right|}{\left|z_{1}\right|} e^{i \theta} \Rightarrow \frac{z_{2} \bar{z}_{1}}{z_{1} \bar{z}}=\frac{\left|z_{2}\right|}{\left|z_{1}\right|} e^{i \theta}\) 
⇒ \(\frac{\mathrm{z}_{2} \overline{\mathrm{z}}_{1}}{\left|\mathrm{z}_{1}\right|^{2}}=\frac{\left|\mathrm{z}_{2}\right|}{\left|\mathrm{z}_{1}\right|} \mathrm{e}^{\mathrm{i} \theta} \Rightarrow \mathrm{z}_{2} \overline{\mathrm{z}}_{1}=\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right| \mathrm{e}^{\mathrm{i} \theta}\)
⇒ \(z_{2} \bar{z}_{1}=\left|z_{1}\right|\left|z_{2}\right|(\cos \theta+i \sin \theta)\)
∴ \(\operatorname{Re}\left(\mathrm{z}_{2} \overline{\mathrm{z}}_{1}\right)=\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right| \cos \theta\) ... (i)
and \(\operatorname{Im}\left(\mathrm{z}_{2} \overline{\mathrm{z}}_{1}\right)=\left|\mathrm{z}_{1}\right|\left|\mathrm{z}_{2}\right| \sin \theta\) ... (ii)
The dot product of z₁ and z₂ is defined by z₁ . z₂ = |z₁| |z₂|
cos θ = Re \(\left(\mathrm{z}_{2} \overline{\mathrm{z}}_{1}\right)\)[from (i)] and cross product of z₁ and z₂ is defined z₁ × z₂ = |z₁| |z₂| sin θ = Im \(\left(\mathrm{z}_{2} \overline{\mathrm{z}}_{1}\right)\) [from Eq. (ii)]
If z₁ = 5 + 12i and z₂ = 3 + 4i, then (the projection of
z₁ on z₂ + projection of z₂ on z₁) is equal to