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If black hole has a charge, they will lose charge due to Hawking radiation. If black hole has positive charge, it emit more positron than electron. And the same argument I can apply in rotating black hole. It sould be lose angular momentum due to Hawking radiation Is it right? How I can calculate it?

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If black hole has a charge, they will lose charge due to Hawking radiation.

This is not correct. Black holes lose energy due to Hawking radiation,not charge. Have a look at my answer here.

If black hole has positive charge,

yes, there are charged black holes considered.

it emit more positron than electron.

The mechanism, Feynman diagrams, of Hawking radiation exist outside the horizon of the black hole. The charges of the black hole are within the horizon most of them at the classical singularity, some still falling in. These charges cannot get out of the black hole. Hawking radiation has equal probability of releasing a positive or negative particle, and thus cannot change the charge of a black hole,( by absorbing more pluses than minuses , for example).

And the same argument I can apply in rotating black hole.It sould be lose angular momentum due to Hawking radiatio

The argument cannot hold with hawking radiation, because the probabilities of being emitted at the horizon are balanced randomly in angular momentum. Unless one builds a very specific quantum mechanical model with the gravitons which interact to emit the hawking radiation.

How I can calculate it?

You would need to think up a successful model which would give higher probability for hawking radiation to happen in one direction of the rotation of the black hole than in the other.

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The entropy of a static black hole is given by Wald's entropy which is derived from Noether charge. Namely $$S_{W}=2\pi\oint_{\mathrm{H}}Q_{\zeta}$$ where $\mathrm{H}$ is horizon and $\zeta$ is the Killing vector and $\zeta=0$ on the horizon.

Now computing the Wald's entropy for the Einstein-Hilbert action gives the famous law for BH entropy which is given by $$S_W=A(4G)^{-1}$$ where A is the area of the horizon. Now if you apply the same rule to the rotating BHs you will get the same area law plus some corrections. In order to do so you need to work in string framework to find the entropy for an extremal BH (ie rotating BH). Look at this for example: http://arxiv.org/pdf/0901.0931v2.pdf

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    $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0901.0931 $\endgroup$ – Qmechanic Mar 9 '15 at 23:42

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