# So easy inequality in complex numbers

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.

• Not true at all, this "so easy inequality", that's why you are struggling. – Teresa Lisbon Apr 5 '19 at 9:06
• You have probably a typo. Instead of $\sqrt 2$ put $2$. – user555729 Apr 5 '19 at 9:09
• @астонвіллаолофмэллбэрг could you please glance my comment below? – user519955 Apr 5 '19 at 9:10

## 2 Answers

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

• OK. Thanks. How can I show |z-i|<sqrt2 when finding taylor expansion around i of 1/z via using Maclaurin expansion of 1/z? – user519955 Apr 5 '19 at 9:09
• @user519955 You cannot show $|z-i| < \sqrt 2$ because it is not true, right? Please clarify your context further, maybe we can infer the right question from this. – Teresa Lisbon Apr 5 '19 at 9:31

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