1) Express the complex number
1 / [1 - (cos θ + i sin θ )] in the cartesian form.
2) Prove that [(1 + cos θ + i sin θ) / (1+cos θ)] n = cos n θ + i sin n θ.
3)(a)If c2+s2=1 ,then evaluate (1 + c + i s) / (1+c- i s)
(b) If Z = r e i θ then find |e iz |.
13) a) let c=cosx , s=sinx.
(1+c+is) / (1+c-is) = (1 + cosx + i sinx) / ( 1 + cosx - i sinx)
=cos x/2 ( cos x/2 + i sin x/2 ) / [ cos x/2 ( cos x/2 - i sin x/2 ) ]
= ( cos x/2 + i sin x/2 )2 = ( cos x + i sin x) = c + is
1Q1,2,3 can be solved by just conjugating,i.e multiplying the numerator and denominator by its conjugate.....
what i found intresting is question 4i might be doing a mistake but lets see
seeking at the question it seemed to be that the e term has ab imaginary angle,
62(b) If Z = r eiθ =a+ib where a and b are real numbers ......
then |e iz | = ea+ib = ea.eib
modulus of this is ea
If Z = r e i θ = r cos θ +i r sin θ
iZ =i r cos θ - r sin θ
|e iz | = e-r sin θ
