# What is the real part of the complex number (1-cosx-i*sinx)/(1+cosx+i*sinx)?

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### 1 Answer

We'll re-write the numerator and denominator of the given complex number, using the half angle identities:

(1-cosx-i*sinx)/(1+cosx+i*sinx) = {2[sin (x/2)]^2 - i*sin x}/{2[cos (x/2)]^2 + i*sin x}

We'll re-write the term sin x, using the double angle identity:

sin x = 2 sin(x/2)*cos (x/2)

(1-cosx-i*sinx)/(1+cosx+i*sinx) = {2[sin (x/2)]^2 - i*2 sin(x/2)*cos (x/2)}/{2[cos (x/2)]^2 + i*2 sin(x/2)*cos (x/2)}

We'll factorize and we'll get:

(1-cosx-i*sinx)/(1+cosx+i*sinx) = 2 sin(x/2){[sin (x/2)] - i*cos (x/2)}/2cos (x/2){[cos (x/2)] + i*2 sin(x/2)}

We'll simplify and we'll get:

(1-cosx-i*sinx)/(1+cosx+i*sinx) = 2 sin(x/2){[sin (x/2)] - i*cos (x/2)}/2cos (x/2){[cos (x/2)] + i*2 sin(x/2)}

(1-cosx-i*sinx)/(1+cosx+i*sinx) = 2 sin(x/2){[cos (x/2)] + i*sin (x/2)}/2cos (x/2)*i*{[cos (x/2)] + i*2 sin(x/2)}

(1-cosx-i*sinx)/(1+cosx+i*sinx) = tan(x/2)/i

Since it is not allowed to keep a complex number at denominator, we'll multiply by i both numerator and denominator.

(1-cosx-i*sinx)/(1+cosx+i*sinx) = i*tan(x/2)/i^2

But i^2 = -1

(1-cosx-i*sinx)/(1+cosx+i*sinx) = -i*tan(x/2)

The rectangular form of a complex number is:

z = x + i*y, where Re(z) = x and Im(z) = y

Comparing both forms, we'll get Re(z) = 0 and Im(z) = - tan(x/2).

**Therefore, the real part of the given complex number is Re(z) = 0.**