Complex numbers problem: $ |\frac 1z - \frac 14 | = \frac 14 $

Question

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).

Answer 1

You are given the modulus of a complex number, $\frac14$. So the argument is unknown and the number has the form $\frac{e^{i\theta}}4$.

$$\frac1z-\frac14=\frac{e^{i\theta}}4,$$ so that $$z=\frac4{1+e^{i\theta}}=\frac4{1+\cos\theta+i\sin\theta}=4\frac{1+\cos\theta-i\sin\theta}{(1+\cos\theta)^2+\sin^2\theta}=2\frac{1+\cos\theta-i\sin\theta}{1+\cos\theta}.$$

As you see that the real part equals $2$ and the imaginary part can take any value.

Answer 2

You can multiply both sides by 4 then use the properties of $|a|=|-a|$ to get that simplification. Then you have to solve $(a-4)^2+b^2=a^2+b^2$, hopefully you can do that but probably not so

$$(a-4)^2+b^2=a^2-8a+16+b^2=a^2+b^2$$ so $a=2$ then solve for $b$

Answer 3

After the hint, you get $|z-4|=|z|$. So you are looking for complex numbers whose distance to $0$ is equal to the distance to $4$. These form the vertical line with real part $2$.

Answer 4

Consider $$\Big|\frac 1z - \frac 14 \Big| = \frac 14.$$ This is equivalent to $$|z-4|=|z|.$$ Squaring both sides, we can obtain $$\overline{(z-4)}(z-4)=|z|^2$$ $$|z|^2-4(z+\bar{z})+16=|z|^2.$$ This implies $$\dfrac{z+\bar{z}}{2}=\Re{z}=2.$$ Therefore $$z=2+iy$$ where $y$ can be any real value.

Answer 5

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.