Show that if $z$ is any point on the line joining $z_1 = 1$ and $z_2 = i$ then $|z|\geq \frac{1}{\sqrt{2}}$

Question

Show that if $z$ is any point on the line joining $z_1 = 1$ and $z_2 = i$ then $|z|\geq \frac{1}{\sqrt{2}}$

Let $z=x+iy$ is any point on the line joining $z_1 = 1$ and $z_2 = i$ i.e on the line $y=1-x$

Then $$|z|=\sqrt{x^2+y^2}=\sqrt{2}\sqrt{x^2-x+1/2}$$

What to do next to get the desired result.

Answer 1

We have

$$|z|^2=x^2+(1-x)^2 \ge 1/2 \iff x^2-x+1/4 \ge 0 \iff (x-1/2)^2 \ge 0.$$

Answer 2

Perpendicular distance of the line from origin is $1/\sqrt 2$. So any point will have distance greater or equal to $1/\sqrt 2$.

Answer 3

Following your approach you only need to find a sharp lower bound for your expression for $|z|$ which can be easily found by square completion:

$$|z|=\sqrt{2}\sqrt{x^2-x+1/2}=\sqrt{2}\sqrt{\left(x-\frac 12\right)^2+\frac 14}\geq \frac{\sqrt{2}}{2} =\frac 1{\sqrt 2}$$

So, you can also see, that the minimum is achieved for $x=\frac 12 \Rightarrow y=\frac 12$.