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Problem: Find all complex numbers $z$ such that $|z|=\frac{1}{|z|}=|1-z|$.

Basically I have an idea how to solve this and I get $x=\frac12$ but how should I express it mathematically? Should I go and find $y$ also?

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    What does $|z| = 1/|z|$ tell you about $|z|$? That's a good first step. (Go ahead and show what you get by working out this part, by editing your original post. That'll show us what you know how to do, and make our answers more useful to you.) – 2017-01-29
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    When: $$\left|\text{z}\right|=\frac{1}{\left|\text{z}\right|}=\left(\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]\right)^\frac{1}{2}=\left(\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]\right)^{-\frac{1}{2}}$$ – 2017-01-29
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    If you had told us which reasoning led you to $x=\frac12$, we could have helped you to formulate it rigorously. – 2017-01-29

3 Answers 3

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$$|z|=\frac1{|z|}\Rightarrow |z|=1,$$ $$|z|=|1-z|\Rightarrow \sqrt{x^2+y^2}=\sqrt{(1-x)^2+y^2}\Rightarrow x=\frac12,$$ where $z=x+iy$. Next, we will determine the imaginary part $$\sqrt{\left(\frac12\right)^2+y^2}=1\Rightarrow y=\pm \frac{\sqrt{3}}2.$$ Therefore solutions are $$z_1=\frac12+i\frac{\sqrt{3}}2,$$ $$z_2=\frac12-i\frac{\sqrt{3}}2.$$

3

Hint: $\require{cancel}$ $|z|^2=|1-z|^2$ $\iff$ $z \bar z = (1-z)(1-\bar z)$ $\iff$ $\cancel{z \bar z}=1-z-\bar z+\cancel{z \bar z}\,$

Since $1=|z|^2=z \bar z$ it follows that $\bar z = \cfrac{1}{z}$ so the last equality becomes $z^2-z+1=0\,$, which is a simple quadratic having the two possible values of $z\,$ as roots.

2

Hint: $|z|=1$ represents a circle in the complex plane where the x-axis is the real part and y-axis is the imaginary part of the complex number(all points are equidistant from the origin.)
And $|z|=|1-z|$ represents the line $x=\frac{1}{2}$ (set of all points equidistant from 0 and 1 should be the perpendicular bisector of the segment joining 1 and 0). Now you are looking for intersection of the two curves.