Wednesday, October 4, 2017

complex numbers - How to compute $sqrt{i + 1}$











I'm currently playing with complex numbers and I realized that I don't understand how to compute $\sqrt{i + 1}$. My math program, sage, returns




$$1.09868411346781 + 0.455089860562227*i$$



How does sage compute that number? I don't even see how one could rewrite $\sqrt{i + 1}$ in a number of the form $a+bi$.


Answer



$i + 1 = \sqrt2 \left ( {1 \over \sqrt 2} + {1 \over \sqrt 2} i\right ) \\
= \sqrt 2 \left( \cos \left( \pi \over 4 \right) + i \sin \left( \pi \over 4 \right) \right ) \\
= \sqrt 2 e^{i \pi \over 4}$



$\sqrt{i +1} =\left( 2\right)^{1 \over 4}e^{i \pi \over 8} = \left( 2\right)^{1 \over 4}\left( \cos \left( \pi \over 8 \right) + i \sin \left( \pi \over 8 \right)\right)$




Well, this is how Wolframalpha calculates.
The other root would be $\left( 2\right)^{1 \over 4}\left( \cos \left( 9\pi \over 8 \right) + i \sin \left(9 \pi \over 8 \right)\right)$


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...