<s>[INST] Complex numbers problem: $ |\frac 1z - \frac 14 | = \frac 14 $ How do you go about solving the following equation $ |\frac 1z - \frac 14 | = \frac 14 $ where $ z = a + bi $.
A hint is provided, and apprently the equation can be simplified to $ | {z-4\over z} | = 1 $ (don't understand how they did this).
 [/INST] We have:
$\left\lvert\dfrac1z-\dfrac14\right\rvert=\dfrac14$. Combine fractions:
$\left\lvert\dfrac{4-z}{4z}\right\rvert=\dfrac14$. Get rid of fractions (multiply by $\lvert4z\rvert$):
$\lvert4-z\rvert=\lvert z\rvert$. Let $z=a+bi$:
$\lvert4-a-bi\rvert=\lvert a+bi\rvert$.
$\sqrt{(4-a)^2+b^2}=\sqrt{a^2+b^2}$. Get rid of square root signs (they're ugly):
$(4-a)^2+b^2=a^2+b^2$. Subtract $b^2$:
$(4-a)^2=a^2$
Now, you can solve and get $a=2$. Thus, $a$ is $2$, and $b$ can be anything.
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