# Transform the following complex numbers in polar form for -π/2<θ<π/2. cosθ-isinθ / 1-cosθ-isinθ

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Student Comments

neela | Student

(For typing convenience we write x for theta).

Hope the given complex number is Z = (cosx-i simx)/(1-cosx-isinx) and not cosx-isinx/1-cosx-isinx.

Z = (cosx-isinx)(1-cosx+isinx)/{(1-cosxisinx)(1-cosx+isinx)}= [(cosx-cos^2x+ sin^2x+i(2sinxcosx-sinx)]/{ (1-cosx)^2+sin^2x}= (cosx-cos^2x+1-cos^2x)+i*sinx(cosx-1)}/{ 1-2cosx+cos^2x+sin^2x}= {1+cosx-2cos^2+isinx(cosx-1)}/{2(1-cosx)}= {(1-cosx)(1+2cosx)+isinx(cosx-1)/{2(1-cosx)}. Threfore,R = (1+2cosx) /2. is the real partand ,I = sinx(cosx-1)/{2(1-cosx)}Therefore the modulus r and arument X in the polar form is given by:r = sqrt(R^2+ I^2} andX = arc tan (I/R).