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
 A: Elaborating slightly on Dr. MV's alternate approach:
We have
$$
8+6i = (a+bi)^2 = (a^2-b^2)+2abi
$$
where $a$ and $b$ are both reals.  Thus,
$$
a^2-b^2 = 8
$$
and
$$
2ab = 6
$$
From the second equation, we can write $ab = 3$, and therefore
$$
b = \frac{3}{a}
$$
If we substitute this into the first equation, we get
$$
a^2 - \frac{9}{a^2} = 8
$$
or, after multiplying both sides by $a^2$ and letting $u = a^2$, we have
$$
u^2-8u-9 = 0
$$
which is an ordinary quadratic equation that can be solved using the quadratic formula, or by simple factoring into $(u+1)(u-9) = 0$.  Remember, however, that as $u = a^2$, it cannot be negative, so we must discard the solution $u = -1$, leaving $u = a^2 = 9$, or $a = \pm 3$.  That leaves $b = \pm 1$ (that is, it is negative whenever $a$ is negative, and positive whenever $a$ is positive).
ETA: Fixed the sign of $b$.  Don't know what I was thinking.
A: METHODOLOGY $1$:  Using Polar Coordinate Conversion

In THIS ANSWER, I used polar coordinate conversion to show that the square root of a complex number can be expressed in rectangular form by
$$\begin{align}
\sqrt{x+iy}=\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}\pm i\,\text{sgn}(y)\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\tag 1
\end{align}$$

Simply use $x=8$ and $y=6$ in $(1)$ to arrive at 
$$\sqrt{8+i6}=\pm (3+i)$$

METHODOLOGY $2$:  Using Strictly Rectangular Coordinates
As an alternative development, we denote $a+ib=\sqrt{x+iy}$.  Upon squaring, we find that
$$a^2-b^2=x \tag 2$$
and
$$2ab=y \tag 3$$
Solving $(2)$ and $(3)$ simultaneously, we find
$$\begin{align}
a&=\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}\\\\
b&=\pm \text{sgn}(y)\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}
\end{align}$$
in agreement with $(1)$
A: Try $3+i$ (Square it out using foil and $i^2=-1$). Note $-3-i$ also works.
A: Hints:
$$8+6i = 10 \exp\left(i\arctan\frac{3}{4}\right)\tag{1} $$
$$ \arctan(x)=2\arctan y\quad\Longrightarrow\quad x = \frac{2y}{1-y^2} \tag{2}$$
and $(x,y)=\left(\frac{3}{4},-3\right)$ is a solution of $(2)$.
A: do you know DeMoivre's theorem?
If $z$ is a complex number in polar form $z = \rho (\cos \theta + i \sin \theta)$ then $z^n = \rho^n (\cos n\theta + i \sin n\theta)$
In this case:
$\rho = 10\\
\theta = \cos^{-1}\frac 45\\
n=\frac12\\
z = 10(\cos \cos^{-1} \frac 45 + i \sin \cos^{-1} \frac45)$  
$\sqrt {z} = z^\frac12 = \sqrt{10}(\cos \frac12\cos^{-1} \frac 45 + i \sin \frac12\cos^{-1} \frac45) $
$\cos \frac12\cos^{-1} \frac 45 = \sqrt{\frac{1+0.8}{2}} = \sqrt{0.9} =\sqrt{\frac 9{10}} \\
\sin \frac12\cos^{-1} \frac 45 = \sqrt{\frac{1-0.8}{2}} = \sqrt{0.1} = \sqrt{\frac1{10}}$
$\sqrt {z} = $$\sqrt{10}(\sqrt{\frac 9{10}} + i \sqrt{\frac 1{10}})\\
3 + i$
A: $$
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: $(3+i)^2 = 8 + 6i$.
A: If $(a+bi)^2=8+6i$, then $a^2+b^2=|8+6i|=2\cdot|4+3i|=2\cdot 5$.
So, let's try $a+bi=zw$ with $|z|=2$ and $|w|=5$. These are easy to find: $z=1+i$, $w=2+i$. But $(1+i)(2+i)=-8-6i$, so we take instead $iz=-1+i$ and $iw=-1+2i$, which gives $a+bi=(iz)(iw)=(-1+i)(-1+2i)=-1-3i$.
