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How can I calculate the exact value of something like that: $|e^{\sqrt{i}}|$

Jim
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jaranna
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    First you need to be able to calculate $\sqrt{i}$, do you know how to do that? – Jim Feb 01 '14 at 18:50
  • I think that it would be: $ i^{\frac{1}{2}} = a \ \frac{1}{2}\ln i = \ln a $ and then I change it to the exponential: $ \frac{1}{2} \ln e^{\frac{\pi i}{2}} = \ln a \ \frac{\pi i }{4} = \ln a \ e^{\frac{\pi i}{4}} = a $ - is it correct? – jaranna Feb 01 '14 at 18:55
  • When I say calculate $\sqrt{i}$ I mean write it as $a + bi$ for some real numbers $a, b$. If you can do that then you can use Euler's formula to calculate $e^{a + bi}$. – Jim Feb 01 '14 at 18:57

3 Answers3

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Assuming principal values of the square root (complex) function:

$$i=e^{\frac\pi2i}\implies \sqrt i=e^{\frac\pi4i}=\frac1{\sqrt2}\left(1+i\right)\implies$$

$$\left|e^{\sqrt i}\right|=\left|e^{\frac1{\sqrt2}(1+i)}\right|=e^{\frac1{\sqrt2}}$$

Do something similar as above if you want the other square root of $\;i\;$ ...

DonAntonio
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  • Thanks! But just one more thing, how do you count this: $|e^{\frac{1}{\sqrt{2}}(1+i)}|$ – jaranna Feb 01 '14 at 20:42
  • As usual, @jaranna: $$e^{\frac1{\sqrt2}(1+i)}=e^{\frac1{\sqrt2}}\left(\cos\frac1{\sqrt2}+i\sin\frac1{\sqrt2}\right)\implies$$

    $$\left|e^{\frac1{\sqrt2}(1+i)}\right|=e^{\frac1{\sqrt2}}\left(\cos^2\frac1{\sqrt{ 2}}+\sin^2\frac1{\sqrt 2}\right)^{1/2}=e^{\frac1{\sqrt2}}$$

     – DonAntonio Feb 02 '14 at 05:12
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Two things :

1) For $z \in \mathbb{C}, \; \vert e^{z} \vert = e^{\Re(z)}$ where $\Re(z)$ denotes the real part of $z$.

2)

$$ \Big( e^{i\frac{\pi}{4}} \Big)^{2} = \Big( -e^{i\frac{\pi}{4}} \Big)^{2} = i $$

So, depending on the root of $i$ you choose, the result is either $\exp(\frac{1}{\sqrt{2}})$ or $\exp(-\frac{1}{\sqrt{2}})$.

pitchounet
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In exponential form $i = e^{i \pi/2}$. Hence $\sqrt{i} = e^{i \pi/4} = \cos \pi/4 + i \sin \pi/4$. Therefore, $e^{\sqrt{i}} = e^{\cos \pi/4 + i \sin \pi/4} = e^{1/ \sqrt{2}}.e^{i/ \sqrt{2}}$. So, you get $|e^{\sqrt{i}}| = e^{1/ \sqrt{2}}$.

Sandeep Thilakan
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