Suppose we have a complex number $z$ such that $|z|=1$ and $$|\frac{z}{z'} + \frac{z'}{z}|=1$$ where $z'$ is conjugate . How many complex number satisfy this? So I simplified second condition as $$\frac{zz + z'z'}{zz'}=\frac{z^2 + z'^2}{|z|^2} $$
Now we may represent any complex no as $z=x+iy$ so putting it in above equation we get $$x^2 - y^2 = \frac{1}{2} $$ but this gives me locus , implying infinitely many point but my answer is incorrect , what's wrong??
On
If $|z| = 1$ then $z \bar{z} = 1$. Also, if $|{z \over \bar{z} } + {\bar{z} \over z} | = 1$, then multiplying by $z \bar{z}$ gives $|z^2 + \bar{z^2}| = 1$.
Since $\operatorname{re} w = {1 \over 2} (w + \bar{w})$, this becomes $2 |\operatorname{re} z^2 | = 1$. Letting $z = e^{i \theta}$, this reduces to $|\cos (2 \theta) | = {1 \over 2}$.
Solving gives $\theta \in \{ \pm { \pi \over 6 }, \pm { \pi \over 3 }\} $, so $z = { \sqrt{3} \over 2} \pm {1 \over 2}i$ or $z = {1 \over 2} \pm { \sqrt{3} \over 2} i $.
On
$$\begin{cases}|z|=1\\{}\\\left|\frac z{\overline z}+\frac{\overline z}z\right|=1\end{cases}\;\;\implies\frac{|z+\overline z|}{|z|^2}=1\implies\left|z+\frac1z\right|=1\iff |z^2+1|=|z|=1$$
If now you write $\;z=\cos t+i\sin t\;$ , we get
$$1=|z^2+1|^2=|\cos^2t-\sin^2t+2i\cos t\sin t+1|^2=$$
$$=\color{red}{\cos^4t-2\cos^2t\sin^2t+\sin^4t}+2\cos^2t-2\sin^2t+1+\color{red}{4\cos^2t\sin^2t}=$$
$$=\color{red}{(\cos^2t+\sin^2t)^2}+2\cos^2t-2\sin^2t+1=2(\cos^2+1-\sin^2t)=2\cos^2t$$
Complete the answer.
Be careful. Assuming that $|z|=1$ and putting $z = x + \mathrm{i}y$ gives $$\left|\frac{z}{z'}+\frac{z'}{z}\right| = 1 \iff \left|x^2-y^2\right|=\frac{1}{2}$$ For any real number, say $\lambda$, we have $|\lambda|=\lambda$ if $\lambda \ge 0$ and $|\lambda| = -\lambda$ if $\lambda < 0$. $$\begin{eqnarray*} \left|x^2-y^2\right| = \frac{1}{2} \iff \left\{ \begin{array}{ccc} x^2-y^2 = \frac{1}{2} & \text{if} & x^2-y^2 \ge 0 \\ \\ x^2-y^2 = - \frac{1}{2} &\text{if}& x^2-y^2 < 0 \end{array}\right. \end{eqnarray*}$$ Your condition that $|z|=1$ also translates into $x^2+y^2=1$.
When $x^2-y^2 \ge 0$, then the solutions to your problem are given by simultaneously solving $$\begin{eqnarray*} x^2-y^2 &=& \frac{1}{2} \\ \\ x^2+y^2 &=& 1 \end{eqnarray*}$$
When $x^2-y^2 < 0$, then the solutions to your problem are given by simultaneously solving $$\begin{eqnarray*} x^2-y^2 &=& -\frac{1}{2} \\ \\ x^2+y^2 &=& 1 \end{eqnarray*}$$
In both cases, you have to solve $x^2-y^2=\pm\frac{1}{2}$ and $x^2+y^2=1$, i.e. find the intersection of a rectangular hyperbola and a circe. The solutions will be a finite number of discrete points. You'll find that there are actually eight solutions.