I want to find the real part of the square root of a complex number $z$
$\operatorname{Re}(\sqrt(z))$
In general I am wondering if the following is true:
$\operatorname{Re}(\sqrt(z))$ = $\sqrt(\operatorname{Re}(z))$
Thanks
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Try with
$$z=2+i\to z^2=3+4i.$$
Do we have
$$2=\sqrt3\ ?$$
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As pointed out in another answer, $\Re\sqrt{z} \ne \sqrt{\Re z}$ for general $z$.
In general, for $z = x + iy = re^{i\theta}$ with $r \ge 0$ and $\theta \in (-\pi,\pi]$, the principal square root of $z$ is defined to be the number $\sqrt{r}e^{i\frac{\theta}{2}}$. This means
$$\Re\sqrt{z} = \sqrt{r}\cos\frac{\theta}{2}$$ If you apply the half angle formula for cosine and use the fact $$\frac{\theta}{2} \in\left (-\frac{\pi}{2},\frac{\pi}{2}\right)\implies \cos\frac{\theta}{2} \ge 0$$ You will find $\cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}}$ and hence
$$\Re\sqrt{z} = \sqrt{\frac{r + x}{2}} = \sqrt{\frac{\sqrt{x^2+y^2} +x }{2}}$$ By a similar argument, you will find $$|\Im \sqrt{z}| = \sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}$$ The sign of $\Im\sqrt{z}$ will be the same as the sign of $\theta$ and hence $y$. As a result, the principal square root is given by following formula:
$$\sqrt{z} = \sqrt{\frac{\sqrt{x^2+y^2} + x}{2}} + \verb/sign/(y)\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}} i$$ where $\verb/sign/(y) = +1, 0, -1$ depends on whether $y$ is positive, zero or negative.
achille hui
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1Hasn’t everything about complex square roots been said before? E.g. in https://math.stackexchange.com/q/44406/42969, which was pointed out in a comment to the answer. – Martin R Nov 14 '20 at 14:15
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@MartinR I do have read most of the answers of that question before answer this. – achille hui Nov 14 '20 at 14:20
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@GCab: It seems to me that there is already some repetition: https://math.stackexchange.com/questions/linked/44406 – Martin R Nov 14 '20 at 14:39
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Let $\sqrt{z}=x+iy$, and suppose that $\Re(\sqrt z)=\sqrt{\Re(z)}$, then $$x=\sqrt{\Re(x^2-y^2+i2xy)}=\sqrt{x^2-y^2}$$
Thus it holds if and only if $\sqrt{z}\in\Bbb R_{\geq 0}$.
cansomeonehelpmeout
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While this may have been covered in other responses, the question and responses here are concise. The answer is easily understood and has rather important general implications. The responses linked in the first comment while elegant and certainly complete do not provide direct answers. I think the responses here that propose trying a simple example are quite useful and do provoke thought without diving into a long proof. Concise solutions to direct questions with large implications should be celebrated not deemed too trivial.
– Kyle Nov 16 '20 at 23:37