Number of complex number z, such that |z|=1and zz+zz=1 is

Let z=x+iy∵ |z|=1 .....(i)∴ x2+y2=1 and zz¯+zz=1⇒ x+iyx−iy+x−iyx+iy=1⇒ (x+iy)2+(x−iy)2x2+y2=1⇒ 2x2−y21=1 from Eq.(i)⇒ x2−y2=±12 ....(ii)From Eqs.(i) and (ii), we get 2x2=1±12=12,32⇒ x2=14,34⇒x=±12,±32 For x=12,y=±32 from Eq.(i) For x=−12,y=±32 from Eq.(i) For x=32,y=±12 from Eq. (i) For x=−32,y=±12 from Eq.(i) ∴ Solutions are 12±i32,−12±i32,32±i2,−32±i2Hence, number of solutions is 8.

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Number of complex number z, such that $| z | = 1$ and $|\frac{z}{z} + \frac{z}{z}| = 1$ is

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answer is 8.

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Detailed Solution

Let $z = x + iy$
$\begin{array}{l} ∵ | z | = 1 . . . . . (i) \\ ∴ x^{2} + y^{2} = 1 \end{array}$
and $|\frac{z}{\bar{z}} + \frac{z}{z}| = 1$
$\begin{array}{l} \Rightarrow |\frac{x + iy}{x - iy} + \frac{x - iy}{x + iy}| = 1 \\ \Rightarrow |\frac{(x + iy)^{2} + (x - iy)^{2}}{x^{2} + y^{2}}| = 1 \end{array}$
$\Rightarrow |\frac{2 (x^{2} - y^{2})}{1}| = 1 [from Eq . (i)]$
$\Rightarrow x^{2} - y^{2} = \pm \frac{1}{2} . . . . (ii)$
From Eqs.(i) and (ii), we get
$2 x^{2} = 1 \pm \frac{1}{2} = \frac{1}{2}, \frac{3}{2}$
$\Rightarrow x^{2} = \frac{1}{4}, \frac{3}{4} \Rightarrow x = \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}$
$For x = \frac{1}{2}, y = \pm \frac{\sqrt{3}}{2} [from Eq . (i)] For x = - \frac{1}{2}, y = \pm \frac{\sqrt{3}}{2} [from Eq . (i)]$
$For x = \frac{\sqrt{3}}{2}, y = \pm \frac{1}{2} [from Eq . (i)] For x = - \frac{\sqrt{3}}{2}, y = \pm \frac{1}{2} [from Eq . (i)] ∴ Solutions are \frac{1}{2} \pm \frac{i \sqrt{3}}{2}, - \frac{1}{2} \pm \frac{i \sqrt{3}}{2}, \frac{\sqrt{3}}{2} \pm \frac{i}{2}, - \frac{\sqrt{3}}{2} \pm \frac{i}{2}$
Hence, number of solutions is 8.