Prove if $|z-1| \le \frac{1}{2}$ then $|\frac{z}{|z|} - 1| \le |z - 1|\sqrt2$ for z complex

Question

As the title, problem is to prove if $|z-1| \le \frac{1}{2}$ then $|\frac{z}{|z|} - 1| \le |z - 1|\sqrt2$ for complex $z$.

I noticed using the condition we can get $|\frac{z}{|z|} - 1| \le \frac{\sqrt2}{2}$.

Then can square both sides to get $2(1 - \frac{\operatorname{Re}z}{|z|}$) $\le \frac{1}{2}$.

From here I try further manipulations but am stuck and unable to proceed.

How would you solve this problem?

Answer 1

Let $\arg z=\theta.$

Thus, it's enough to show that: $$|\cos\theta+i\sin\theta-1|\leq\frac{1}{\sqrt2}$$ or $$2-2\cos\theta\leq\frac{1}{2}$$ or $$\cos\theta\geq\frac{3}{4},$$ which is $$0^{\circ}\leq \theta\leq\arccos\frac{3}{4}$$ or $$360^{\circ}-\arccos\frac{3}{4}\leq \theta<360^{\circ},$$ which is true because by the given $$0^{\circ}\leq \theta\leq30^{\circ}$$ or $$330^{\circ}\leq \theta<360^{\circ}.$$

Answer 2

Let $z=re^{i\theta}$, and then you need to prove $|e^{i\theta}-1|^2\leq2|re^{i\theta}- 1|^2$. Here $1/2\leq r\leq 3/2,-\pi/6\leq \theta\leq\pi/6$. $$LHS=2-2\cos\theta;$$ $$RHS=2(r^2-2r\cos\theta+1);$$ $$\frac{1}{2}(RSH-LHS)=r^2-2r\cos\theta+\cos\theta\geq 0.$$