Find $\sqrt{8+6i}$ in the form of $a+bi$

Solution 1:

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.

Solution 2:

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 $(1)$

Solution 3:

Try $3+i$ (Square it out using foil and $i^2=-1$). Note $-3-i$ also works.

Solution 4:

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 $(2)$.

Solution 5:

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$