What is $\Im\left(\frac{1}{\sqrt{b-i}}\right)$

Question

What is: $$\Im\left(\frac{1}{\sqrt{b-i}}\right)$$Where $b$ is a real number.

I tried and couldn't do it. Please answer it if you can.

Edit: I tried to ged rid of the $i$ in denominator by multiplying $\sqrt{\frac{b-i}{b-i}}$ and $\frac{b+i}{b+i}$

Then I got stuck. Dont know how to deal with

$\Im\left((b+i)\left(\sqrt{b-i}\right)\right)$

Answer 1

You're heading in a good direction with your idea of multiplying top and bottom by the same thing to make the bottom nicer (which in this case would mean making it a pure real). And you tried the conjugate (i.e. $b+i$) and using a square root, but I think you'll have the better luck if you use both, i.e. multiply by $$ \dfrac{\sqrt{b+i}}{\sqrt{b+i}}$$

From the statement of your question it looks like you're looking for a 'simplification', and it's not always clear what 'simplification' means. If you need to remove all references to imaginary numbers it'll get messy, and YAlexandrov's comment shows what you can expect.

Answer 2

We obtain \begin{align*} \color{blue}{\Im\left(\frac{1}{\sqrt{b-i}}\right)}&=\Im\left((b-i)^{-\frac{1}{2}}\right)\\ &=\Im\left(\exp\left(-\frac{1}{2}\mathrm{Log}(b-i)\right)\right)\tag{1}\\ &=\Im\left(\exp\left(-\frac{1}{2}\ln|b-i|-\frac{i}{2}\mathrm{Arg}(b-i)\right)\right)\tag{2}\\ &=\Im\left(|b-i|^{-\frac{1}{2}}\exp\left(-\frac{i}{2}\mathrm{Arg}(b-i)\right)\right)\tag{3}\\ &=\Im\left((b^2+1)^{-\frac{1}{4}}\left[\cos\left(\frac{1}{2}\mathrm{Arg}(b-i)\right)-i\sin\left(\frac{1}{2}\mathrm{Arg}(b-i)\right)\right]\right)\tag{4}\\ &\,\,\color{blue}{=-(b^2+1)^{-\frac{1}{4}}\sin\left(\frac{1}{2}\mathrm{Arg}(b-i)\right)}\tag{5}\\ \end{align*}

Comment:

• In (1) we use the representation as Principal value of the complex logarithm.

• In (2) we write the principal value using the real logarithm $\ln$ and the argument $\mathrm{Arg}$.

• In (3) we factor out $|b-i|^{-\frac{1}{2}}$.

• In (4) we write the absolute value $|b-i|=(b^2+1)^{\frac{1}{2}}$ and apply Euler's formula.

• In (5) we select the imaginary part.

Answer 3

Because you're working with a square root, you don't necessarily need to invoke polar form or trigonometric functions: $$\sqrt{u+v\mathrm{i}}=\sqrt{\frac{\sqrt{u^2+v^2}+u}{2}}+\mathrm{i}\,\mathrm{sgn}\,v\sqrt{\frac{\sqrt{u^2+v^2}-u}{2}}\quad(u\neq 0\text{ or } v\geq 0)$$ (Abramowitz and Stegun 3.2.27; when $u^2+v^2=1$, this equality reduces to the half-angle formulae for trigonometric functions.)

Then $$\begin{split}\Im\frac{1}{\sqrt{b-\mathrm{i}}}&=\frac{\Im((b+\mathrm{i})\sqrt{b-\mathrm{i}})}{b^2+1}\\ &=\frac{b\Im\sqrt{b-\mathrm{i}}+\Re\sqrt{b-\mathrm{i}}}{b^2+1}\\ &=\frac{-b\sqrt{2(\sqrt{b^2+1}-b)}+\sqrt{2(\sqrt{b^2+1}+b)}}{2(b^2+1)} \end{split}$$ which is a "simplification" in that trigonometric functions have been eliminated in favor of real square roots.