Q;{I'm stuck on part (iii) }

Given w=1-e^(jx)cosx 0<x<pie/2

(i) express e^jkx and e^-jkx in the form a+jb

(ii) show w=-je^(jx)sinx

(iii)find |W| and argw, hence write down the modulus and argument of each of the two square roots of w.

k so, bare with me.. (ii) is just an expansion of w in original Q giving on the way w=-jsinx(cosx+jsinx) FROM here i would conclude that |w| is

|-jsinx| = |sinx| = sinx (correct) ,, my teacher said do,|-j|.|e^(jx)|.|sinx| {more complicated i think} so that's ok.

Looking at w=-jsinx(cosx+jsinx) i would therefore think that argument is just x,,,,,,,as this is wrong ,presumably the modulus of sinx has some interplay into the argument 'i dunno'. So is it reasonable to conclude everytime a modulus is not a whole no?????? i must write arg(-j)arg(e^jx)arg(sinx), plot each one and get x-pie/2,,,,i presume so,, don't know y,, n e way not to worry,,

|(w)^o.5| = (sinx)^0.5,,,so argument will be using de moivre (x-pie/2)/2,,,,,,,,,,,,,,and the other one is{don't know where came from}

((2x-pie)/4)+pie i think it has something to do with the limits but i can't see how!

If n e one could take the time to help me on part (iii) that would be most appreciated thnx