$|z|=1$ should represent semi-circle or circle? Suppose we have complex number $z=x+iy$ and we are given locus $|z|=1$ which should be $\sqrt{x^2+y^2} =1$ this should be a semi-circle above x axis , it's when we square our equation we get a circle (given condition only points towards a semi-circle). So, why do people call locus a circle ?
 A: It's possible that squaring an equation can "double up" the solutions to include negatives that they didn't before. For instance, the graph of $y=\sqrt{x}$ (for $x>0$) is just a single component that exists above the $x$-axis, whereas its square $y^2=x$ (again with $x>0$) would the same component above the $x$-axis as well as its mirror image below the $x$-axis.
However there is no rule that says squaring an equation doubles up the components of the solution set. For instance, squaring our $y^2=x$ to get $y^4=x^2$ yields no new solutions. Even if the original equation involved radicals, squaring it may not yield new solutions. Indeed, what goes on when solution sets expand is that the original radicand is newly allowed to be negative where it wasn't before, but in our case the radicand $x^2+y^2$ still can't be negative even with the square root removed so there are no new solutions going from $\sqrt{x^2+y^2}=1$ to $x^2+y^2=1$.
When $a\ge0$, we know $\sqrt{a}=b$ and $a=b^2$ are logically equivalent, so we can interchange them on our whim. Since $x^2+y^2\ge0$ always, $\sqrt{x^2+y^2}=1$ is always equivalent to $x^2+y^2=1$, so any point that is a solution to one must always be a solution to the other. Check for yourself: plug in any point on the unit circle below the $x$-axis into $\sqrt{x^2+y^2}=1$ and see what happens.
This shouldn't be a surprise. Since the modulus $\sqrt{x^2+y^2}$ (by the Pythagorean theorem) is the distance from $(0,0)$ to the point $(x,y)$, the equation $\sqrt{x^2+y^2}=1$ describes all points exactly one unit of distance from the origin. That's the entire unit circle.
Indeed, it's a bit arbitrary that you'd guess the solution set is the part of the unit circle with nonnegative $y$-values. Perhaps parallel-universe you would ask this same question on parallel-universe MSE, but instead claim the solution set is the part of the unit circle with nonnegative $x$-values. Maybe the reason you're afraid of having two $y$-values for the various $x$-values is that this would not be the graph of a function, but this is no issue. The equation $\sqrt{x^2+y^2}=1$ does not describe a function any more than $x^2+y^2=1$ does.
Now, if you had instead spoke of $y=\sqrt{1-x^2}$, then you'd be talking about the graph of a function, and this would yield the part of the unit circle with nonnegative $y$-values. Then if you squared that equation you'd get $y^2=1-x^2$, which is equivalent to $x^2+y^2=1$, and that includes all solution on the unit circle.
A: Observe that: $|z| =1 \iff \sqrt{x^2+y^2} = 1 \iff x^2+y^2 = 1$, this represents a full circle and not a semi-circle.
A: I believe you're being confused by the example of $y=\sqrt{x}$ which is half of a parabola. 
This is different from $\sqrt{x^2+y^2}=1$ because in the first equation $y$ cannot be negative and in the second equation it can be. 
