<s>[INST] Find $\sqrt{8+6i}$ in the form of $a+bi$ I need help with changing $\sqrt{8+6i}$ into complex number standard form.
I know the basics of complex number such as the value of $i$ and $i^2$, equality of complex number, conjugate and rationalizing method. This is my first encounter with such term and its only been a week since I learn about complex number. I would very appreciate a thorough explanation or just a calculation would suffice
Edit- Thank you guys, I've found both answer. And about the methods that Im not familiar with, It'll probably make sense to me after the next few class. I'll make sure to come back here and study them. Thanks again! You guys are great :D
 [/INST] $$
8+6i = \sqrt{8^2+6^2} \cdot (\cos\theta + i\sin\theta) = 10(\cos\theta+i\sin\theta)
$$
where $\cos\theta = \dfrac 8 {\sqrt{8^2+6^2}} = \dfrac 8 {10} = \dfrac 4 5$ and $\sin\theta = \dfrac 6 {\sqrt{8^2+6^2}} = \dfrac 6 {10} = \dfrac 3 5$. 
The square roots are
$$
\pm\sqrt{10} \left( \cos\frac\theta 2 + i \sin \frac\theta 2 \right).
$$
For the tangent half-angle formula we get
$$
\tan\frac \theta 2 = \frac{\sin\theta}{1+\cos\theta} = \frac{3/5}{1+ 4/5} = \frac 3 {5+4} = \frac 3 9 = \frac 1 3.
$$
So $\cos\dfrac\theta2 = \sqrt{\dfrac 1 {1+\tan^2\frac\theta 2}} = \dfrac 3 {\sqrt{10}}$ and $\sin\dfrac\theta2 = \dfrac1{\sqrt{10}}$.
Thus the square roots are
$$
\pm\sqrt{10} \left( \frac 3 {\sqrt{10}} + i \frac 1 {\sqrt{10}} \right) = \pm(3+i).
$$
It's easy to check this by multiplying: $(3+i)^2 = 8 + 6i$.
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