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Mathematics Question on Complex Numbers and Quadratic Equations

Let a=Im(1+z22iz),a=Im\left(\frac{1+z^{2}}{2iz}\right), where zz is any non-zero complex number. The set A=\left\\{a:\left|z\right|=1\,and\,z\ne\pm1\right\\} is equal to:

A

(−1,1)( - 1, 1)

B

[−1,1][ - 1, 1]

C

[0,1)[0, 1)

D

(−1,0]( - 1, 0]

Answer

(−1,1)( - 1, 1)

Explanation

Solution

The correct answer is (A) : (−1,1)( - 1, 1)
Let z=x+iy⇒z2=x2−y2+2ixyz=x+iy \Rightarrow z^{2}=x^{2}-y^{2}+2ixy
Now,
1+z22iz=1+x2−y2+2ixy2i(x+iy)=(x2−y2+1)+2ixy2ix−2y\frac{1+z^{2}}{2iz}=\frac{1+x^{2}-y^{2}+2ixy}{2i\left(x+iy\right)}=\frac{\left(x^{2}-y^{2}+1\right)+2ixy}{2ix-2y}
=(x2−y2+1)+2ixy−2y+2ix×−2y−2ix−2y−2ix=\frac{\left(x^{2}-y^{2}+1\right)+2ixy}{-2y+2ix}\times \frac{-2y-2ix}{-2y-2ix}
=y(x2−y2−1)+x(x2+y2+1)i2(x2+y2)=\frac{y\left(x^{2}-y^{2}-1\right)+x\left(x^{2}+y^{2}+1\right)i}{2\left(x^{2}+y^{2}\right)}
a=x(x2+y2+1)2(x2+y2)a=\frac{x\left(x^{2}+y^{2}+1\right)}{2\left(x^{2}+y^{2}\right)}
Since, ∣z∣=1\left|z\right|=1
⇒x2+y2=1\Rightarrow \sqrt{x^{2}+y^{2}}=1
⇒x2+y2=1\Rightarrow x^{2}+y^{2}=1
∴a=x(1+1)2×1=x\therefore a=\frac{x\left(1+1\right)}{2\times1}=x
Also z≠1z\ne1
⇒x+iy≠1\Rightarrow x+iy\ne1
∴A=(−1,1)\therefore A=\left(-1,1\right)