For how many real values of $c$ do we have $\left|\frac12-ci\right| = \frac34$?
I recall that the absolute value of a complex number $a+bi$ is the same as the distance from the origin to the point $(a,b)$ on the complex plane. So, the equation $\left|\frac12-ci\right| = \frac34$ means that the distance from $(\frac12, -c)$ to the origin is $\frac34$. I can use the Pythagorean theorem to write this as $\left(\frac12\right)^2 + (-c)^2 = \left(\frac34\right)^2$. Simplifying, I get $\frac14 + c^2 = \frac{9}{16}$. Subtracting $\frac14$ from both sides, I get $c^2 = \frac{5}{16}$. Taking the square root of both sides, I get $c = \pm \frac{\sqrt{5}}{4}$. These are the only two possible values of $c$ that satisfy the equation, since any other value would make the distance larger or smaller than $\frac34$. Therefore, the answer is 2. 