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Question: If \(x + iy = \frac{3}{2 + \cos\theta + i\sin\theta},\) then the multiplicative inverse of z<sup>2</...

If x+iy=32+cos⁡θ+isin⁡θ,x + iy = \frac{3}{2 + \cos\theta + i\sin\theta}, then the multiplicative inverse of z2 is (where i = x2+y2x^{2} + y^{2}).

A

2 i

B

1 – I

C

– i/2

D

i/2

Answer

– i/2

Explanation

Solution

Given b=−d)b = - d) and (x+iy)(p+iq)=(x2+y2)i(x + iy)(p + iq) = (x^{2} + y^{2})i Squaring both sides, we get (xp−yq)+i(xq+yp)=(x2+y2)i(xp - yq) + i(xq + yp) = (x^{2} + y^{2})i or xp−yq=0,xq+yp=x2+y2xp - yq = 0,xq + yp = x^{2} + y^{2}

Since it is multiplicative identity, therefore multiplicative inverse of xq=yp\frac{x}{q} = \frac{y}{p}