Real part of $\sqrt{ai-1}$ 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
$$
i=e^{i\pi/2}=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}.
$$

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...
