Plz post the question clearly. ...
- Sahil Jain its quite clear...nahi aata toh aise bol do.Upvote·0· Reply ·2013-07-29 07:06:01
\mathbf{z_{1}=r_{1}e^{i\theta_{1}},}\mathbf{z_{2}=r_{2}e^{i\theta_{2}}}
Use,
eiθ=cosθ+isinθ
And find z1z2.You'll find that:
\mathbf{Re(z_{1}z_{2})=r_{1}r_{2}(cos(\theta_{1})cos(\theta_{2})-sin(\theta_{1})sin(\theta_{2}))}
=\mathbf{r_{1}cos(\theta_{1})r_{2}cos(\theta_{2})-r_{1}sin(\theta_{1})r_{2}sin(\theta_{2})}
\mathbf{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}
Plz post the question clearly. ...