Show that if $|z|=1$ then $|e^z|>\frac{1}{3}$

Question

i knew that from modulus in complex,

$|e^z|=|e^{x+iy}|\\=|e^x.e^{iy}|\\=|e^x|.|cos(y)+isin(y)|\\=|e^x|.\sqrt{cos^2(y)+sin^2(y)}\\=|e^x|$

and from $$|z|=1\\\sqrt{x^2+y^2}=1\\x^2+y^2=1$$

i don't know what to do from there, how can i prove it?

$e^x \ge e^{-1} > 1/3$, the latter because $e < 3$. – Martin R Apr 21 '21 at 19:41 — Martin R, Apr 21 '21 at 19:41

score 2 · Answer 1 · answered Apr 21 '21 at 19:43

2

$\left| e^{z}\right|=e^x$ is minimized when $x$ is chosen to be the smallest possible value it can, and if $x^2+y^2=1$ then then least value for $x$ is $x=-1$ (corresponding to the point (-1,0)), at which point $\left| e^{z}\right|=e^{-1}\approx0.368>1/3$.

answered Apr 21 '21 at 19:43

Milo Moses

2,517

thankyou so much! – Ceciliads Apr 22 '21 at 02:48

score 1 · Accepted Answer · answered Apr 21 '21 at 19:44

1

You are almost there, basically, you have shown that $|e^z| = e^x$, with $x$ being the real part of $z$, note that $x \in [-1,1]$, so $$e^x \geq e^{-1} \approx 0.3679 > \frac{1}{3}.$$

answered Apr 21 '21 at 19:44

Fei Cao

2,830

thankyou so much! – Ceciliads Apr 22 '21 at 02:48
@Ceciliads No worries! – Fei Cao Apr 22 '21 at 03:43

Show that if $|z|=1$ then $|e^z|>\frac{1}{3}$

2 Answers2