6

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

Afiq Afandi
  • 69

7 Answers7

3

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.

Brian Tung
  • 34,160
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 $(1)$

Community
  • 1
Mark Viola
  • 179,405
1

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

coffeemath
  • 29,884
  • 2
  • 31
  • 52
1

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 $(2)$.

Jack D'Aurizio
  • 353,855
  • @Dr.MV: that is not surpring, since we are dealing with a square root. I was just pointing the OP towards the fact that $1-3i$ (or, better, $\pm(i+3)$) has something to do with the solutions he is looking for. – Jack D'Aurizio Jun 06 '16 at 15:53
  • I doubt the op has learned the polar representation of complex numbers yet. I'd be very surprised if the first line would make any sense to a student who only learned complex numbers existed a week ago. – fleablood Jun 06 '16 at 15:56
  • 2
    @fleablood: it is not in my power to control what the OP knows or does not, but an alternative approach to solving $$ (a+bi)^2 = (a^2-b^2)+i(2ab) = 8+6i$$ might be interesting as a teaching opportunity, I think. – Jack D'Aurizio Jun 06 '16 at 15:59
  • Um, it's within your power to provide an answer that is useful for the OP. What the heck is the point of answering a question here if it isn't to help the OP. No-one cares if we know how to solve this question. The only thing that matters is the OP asked for help that he or she would understand. – fleablood Jun 06 '16 at 16:04
  • @fleablood was right. But I'll come back to this for sure once I've learned more of this topic. Thank you very much – Afiq Afandi Jun 06 '16 at 16:04
  • 2
    "Answers are for everybody." This answer was the third one, it provides a useful approach, and it wasn't exclusionary about it. – Brian Tung Jun 06 '16 at 16:08
  • @AfiqAfandi don't be put off. In the weeks to come you will learn that the complex numbers, having 2 dimensions, have a geometric representation where a complex number has a "size" r = $\sqrt{a^2 + b^2}$ -- its distance from 0, and an angle $\theta= \arccos (b/r)$ from the x axis. It will turn out that a+bi = $r(\sin \theta + \cos \theta i)$ which may seem complicated but once we discover that $e^{\theta i} = \sin \theta + i\cos \theta$ (!!!!!) this will actually become a very useful 2nd way to right complex numbers. – fleablood Jun 06 '16 at 16:18
  • ""Answers are for everybody." This answer was the third one, it provides a useful approach, and it wasn't exclusionary about it." -- fair enough. – fleablood Jun 06 '16 at 16:19
1

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$

Michael Hardy
  • 1
Doug M
  • 57,877
1

$$ 8+6i = \sqrt{8^2+6^2} \cdot (\cos\theta + i\sin\theta) = 10(\cos\theta+i\sin\theta) $$ 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: $(3+i)^2 = 8 + 6i$.

Michael Hardy
  • 1
0

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 $a+bi=(iz)(iw)=(-1+i)(-1+2i)=-1-3i$.

lhf
  • 216,483