So easy inequality in complex numbers

Question

Show that $$|z| \lt 1 \Rightarrow |z-i| \lt \sqrt 2$$ $x^2+y^2 \lt 1$ How can I show $x^2+(y-1)^2 \lt 2$ ? I’m sorry, i know it’s so easy but I couldnt obtain it in no way.

Answer 1

This is incorrect. Consider $z=-\frac{1}{2}i$. Then $|z|=\frac{1}{2}<1$ but $|z-i|=\left|-\frac{1}{2}i-i\right|=\left|-\frac{3}{2}i\right|=\frac{3}{2}>\sqrt{2}$.

Answer 2

After the correction $\sqrt 2\to 2$: use the triangle inequality: $$|z - i|\le |z| + |i|< 1 + 1 = 2.$$