Real part of $\sqrt{ai-1}$

Question

Is there a way to find the real and imaginary parts of $$ z=\sqrt{ai-1},\qquad a>0 $$ where $i=\sqrt{-1}$. Thanks. I do not know what to to do. Note $$ i=e^{i\pi/2}=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}. $$

Answer 1

Convert $ai-1$ into polar form. Let $ai-1 = r(\cos\theta + i\sin\theta)$. Then $$\begin{align*} a &= r\sin\theta\\-1 &= r\cos\theta\\ r^2 &= (r\sin\theta)^2 + (r\cos\theta)^2\\ &= a^2 + 1\\ r &= \sqrt{a^2 + 1}\\ \tan \theta &= \frac{r\sin\theta}{r\cos\theta}\\ &= -a\\ \theta &= \pi - \tan^{-1} a \end{align*}$$ Let $r(\cos\theta + i\sin\theta) = [p(\cos\phi + i\sin\phi)]^2 = p^2(\cos2\phi + i\sin2\phi)$ for $p\ge 0$.

Then $p^2 = r$ and $p = \sqrt r$. Also, there are two arguments $\phi\in[0,2\pi)$ that will satisfy $2\phi\equiv \theta \pmod {2\pi}$: $$\phi_1 = \frac\theta 2,\quad \phi_2 = \frac{\theta + 2\pi}2$$ So one of the roots is $$\begin{align*} p(\cos\phi_1 + i\sin\phi_1) &= \sqrt[4]{a^2+1}\left[\cos\frac{\pi-\tan^{-1}a}2 + i\sin\frac{\pi-\tan^{-1}a}2\right]\\ &= \sqrt[4]{a^2+1}\left[\sin\frac{\tan^{-1}a}2 + i\cos\frac{\tan^{-1}a}2\right]\\ \end{align*}$$ and the other root is $$\begin{align*} p(\cos\phi_2 + i\sin\phi_2) &= \sqrt[4]{a^2+1}\left[\cos\frac{3\pi-\tan^{-1}a}2 + i\sin\frac{3\pi-\tan^{-1}a}2\right]\\ &= \sqrt[4]{a^2+1}\left[-\sin\frac{\tan^{-1}a}2 - i\cos\frac{\tan^{-1}a}2\right]\\ \end{align*}$$ Simplify the nested inverse tangent in (co)sine using half-angle formulae. $$\begin{align*} \sin\frac{\tan^{-1} a}2 &= \operatorname{sgn} a \sqrt{\frac{1-\cos\tan^{-1}a}2}\\ &= \operatorname{sgn} a\sqrt{\frac{1-\frac1{\sqrt{a^2+1}}}2}\\ &= \operatorname{sgn} a\sqrt{\frac{\sqrt{a^2+1} - 1}{2\sqrt{a^2+1}}}\\ \cos\frac{\tan^{-1} a}2 &= \sqrt{\frac{1+\cos\tan^{-1}a}2}\\ &= \sqrt{\frac{1+\frac1{\sqrt{a^2+1}}}2}\\ &= \sqrt{\frac{\sqrt{a^2+1} + 1}{2\sqrt{a^2+1}}} \end{align*}$$ For this question, $\operatorname{sgn} a$ is always $1$.

Answer 2

for $z=x+iy$ you can right its Polar Representation $z=re^{i\theta‎}=r(\cos\theta+i\sin\theta)=\color{red}{r\cos\theta}+i\color{blue}{r\sin\theta}$(respectively real part and imaginary part red and blue) where $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\frac{y}{x} $ for $x>0$ and $\theta=tan^{-1}\frac{y}{x}+\pi $ for $x<0$.

For $w = \rho e^{i \phi}$ that $w^n = z$, $\rho = r^{1/n}$, $\phi = \frac{\theta}{n} + \frac{2\pi k}{n}$ for $k=0, 1, ..,n-1$.

here for $z=ai-1$, we have $r=\sqrt{x^2+y^2}=\sqrt{(-1)^2+a^2}$ and $\theta=tan^{-1}\frac{a}{-1}+\pi $ thus answers will be $w=({a^2+1})^\frac{1}{4}e^{i(\frac{\theta}{2} + \frac{2\pi k}{2})}$(use method in used in red and blue part you will have) real part first answer is $({a^2+1})^\frac{1}{4}cos\frac{\alpha}{2}$ and imaginary part is $({a^2+1})^\frac{1}{4}sin\frac{\alpha}{2}$ where $\alpha=tan^{-1}(-a)+\pi$ and...

Answer 3

In THIS ANSWER, I developed the expression for the square root of a complex number in Cartesian Coordinates. Using that expression for $z=ai-1$ gives

$$\bbox[5px,border:2px solid #C0A000]{z^{1/2}=\pm \left(\sqrt{\frac{\sqrt{1+a^2}-1}{2}}+ i \sqrt{\frac{\sqrt{1+a^2}+1}{2}}\,\,\,\right)}$$