Friday, March 23, 2018

complex numbers - I am getting $i^{-1}=pm i$.


I am trying to find $i^{-1}$. I already know that the answer is $-i$, but I can't figure out a way to determine that using math. This is what I am doing: $$i^{-1}$$ $$\frac1i$$ $$\sqrt{\left(\frac1i\right)^{2}}$$ $$\sqrt{\frac{1}{-1}}$$ $$\sqrt{-1}$$ $$\pm i$$ What am I doing wrong?


Answer



I'm not sure where your 3rd expression comes from. In general, to find $\frac{1}{a+bi}$, where $a,b\in \mathbb{R}$, multiply numerator and denominator by the conjugate $\overline{a+bi}=a-bi$ to obtain $$\frac{1}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2+b^2}$$ In your case, take $a=0,b=1$ to obtain $\frac{1}{i}=\frac{-1i}{1}=-i$.


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...