# Complex Numbers

• Mar 1st 2014, 02:44 PM
Fratricide
Complex Numbers
a) Express -4√3 - 4i in exact polar form.
b) Find the cube roots of
-4√3 - 4i.
c) Show that the cubic equation z3 - 3
√3iz2 - 9z + 3√3i = -4√3 - 4i can be written in the form (z - w)3 = -4√3 - 4i where w is a complex number.

With part c, I know I have to somehow use my answer(s) from part b in order to solve for w, but I'm not sure exactly how. I was thinking along the lines of:
(z - w)3 = -4√3 - 4i
w = z - 3√(-4√3 - 4i)

Do I have to express the cube roots of
-4√3 - 4i from part b in cartesian form? If so, how do I go about it (considering there are 3 solutions)?
• Mar 1st 2014, 02:47 PM
romsek
Re: Complex Numbers
oh boy.. complex roots... I think I'll just sit back and watch the fireworks :D
• Mar 1st 2014, 03:13 PM
Plato
Re: Complex Numbers
Quote:

Originally Posted by Fratricide
a) Express -4√3 - 4i in exact polar form.
b) Find the cube roots of -4√3 - 4i.
c) Show that the cubic equation z3 - 3√3iz2 - 9z + 3√3i = -4√3 - 4i can be written in the form (z - w)3 = -4√3 - 4i where w is a complex number.

With part c, I know I have to somehow use my answer(s) from part b in order to solve for w, but I'm not sure exactly how. I was thinking along the lines of:
(z - w)3 = -4√3 - 4i
w = z - 3√(-4√3 - 4i)
Do I have to express the cube roots of -4√3 - 4i from part b in cartesian form? If so, how do I go about it (considering there are 3 solutions)?

Let $z=-4\sqrt3 -4i$. Note that $\arctan \left( {\frac{{ - 4}}{{ - 4\sqrt 3 }}} \right) = \dfrac{\pi }{6}$

Thus $\text{Arg}(z)=\dfrac{-5\pi}{6}$.

Let $\theta = \dfrac{{Arg(z)}}{4} = \dfrac{{ -5 \pi }}{24}~ , ~\lambda = \sqrt[4]{{\left| z \right|}}\exp (i\theta )$, and $\rho = \exp \left( {\frac{\pi }{2}} \right)$

Now your roots are $\lambda \rho^k,~~k=0,1,2,3$
• Mar 1st 2014, 04:08 PM
Fratricide
Re: Complex Numbers
Quote:

Originally Posted by Plato
Let $z=-4\sqrt3 -4i$. Note that $\arctan \left( {\frac{{ - 4}}{{ - 4\sqrt 3 }}} \right) = \dfrac{\pi }{6}$

Thus $\text{Arg}(z)=\dfrac{-5\pi}{6}$.

Let $\theta = \dfrac{{Arg(z)}}{4} = \dfrac{{ -5 \pi }}{24}~ , ~\lambda = \sqrt[4]{{\left| z \right|}}\exp (i\theta )$, and $\rho = \exp \left( {\frac{\pi }{2}} \right)$

Now your roots are $\lambda \rho^k,~~k=0,1,2,3$

Thanks for the reply, but you lost me at line 3 (specifically "exp(iθ)" and accompanying notation; it's early days in the course).
• Mar 1st 2014, 04:40 PM
romsek
Re: Complex Numbers
Quote:

Originally Posted by Fratricide
Thanks for the reply, but you lost me at line 3 (specifically "exp(iθ)" and accompanying notation; it's early days in the course).

can you do part (a)? If so you get

$r=\left|z\right|$ and $\theta=\arg(z)$ then $z=r e^{\displaystyle \imath \theta}=r\exp(\imath \theta)$
• Mar 1st 2014, 04:49 PM
Plato
Re: Complex Numbers
Quote:

Originally Posted by Fratricide
Thanks for the reply, but you lost me at line 3 (specifically "exp(iθ)" and accompanying notation; it's early days in the course).

It is really sad that you are taught a subject without being given the full story.

$\exp(i\theta)=\cos(\theta)+i\sin(\theta)$
• Mar 1st 2014, 05:57 PM
Fratricide
Re: Complex Numbers
Quote:

Originally Posted by romsek
can you do part (a)? If so you get

$r=\left|z\right|$ and $\theta=\arg(z)$ then $z=r e^{\displaystyle \imath \theta}=r\exp(\imath \theta)$

Yes, I have done parts (a) and (b).

Quote:

Originally Posted by Plato
It is really sad that you are taught a subject without being given the full story.

$\exp(i\theta)=\cos(\theta)+i\sin(\theta)$

Ahhh, I see. Thank you. Could you elaborate more on those last couple of lines, please? Why do we put Arg(z) over 4 and take the 4th root of |z|? And if exp(iθ) = cosθ + isinθ, then does exp(π/2) = cos(π/2) + sin(π/2)?
• Mar 2nd 2014, 04:21 PM
Plato
Re: Complex Numbers
Quote:

Originally Posted by Fratricide
Could you elaborate more on those last couple of lines, please? Why do we put Arg(z) over 4 and take the 4th root of |z|? And if exp(iθ) = cosθ + isinθ, then does exp(π/2) = cos(π/2) + sin(π/2)?

I will give an outline for finding roots. Suppose $w\in\mathbb{C}$ and $n\in\mathbb{Z}^+$.
Now there are $n~n^{th}$ roots of $w$. Here is the outline of how to find them.

First find $r = \sqrt[n]{{\left| w \right|}}$ Next find $\theta=\text{Arg}(w)$, that is the principal argument, i.e. $-\pi<\theta\le\pi$.

Then the principle root is $\rho = r\exp \left( {\frac{\theta }{n}} \right)$ now is it clear to you that $\rho^n=w~?$

Now the $n$ roots are equally spaced out on a circle centered at zero with radius $r$.

So define a "rotation tool": $\xi = \exp \left( {\frac{{2\pi }}{n}} \right)$.

Now list the roots $\rho\cdot\xi^k~,~k=0\cdots n-1$.
• Mar 2nd 2014, 05:37 PM
Prove It
Re: Complex Numbers
Quote:

Originally Posted by Fratricide
a) Express -4√3 - 4i in exact polar form.
b) Find the cube roots of
-4√3 - 4i.
c) Show that the cubic equation z3 - 3
√3iz2 - 9z + 3√3i = -4√3 - 4i can be written in the form (z - w)3 = -4√3 - 4i where w is a complex number.

With part c, I know I have to somehow use my answer(s) from part b in order to solve for w, but I'm not sure exactly how. I was thinking along the lines of:
(z - w)3 = -4√3 - 4i
w = z - 3√(-4√3 - 4i)

Do I have to express the cube roots of
-4√3 - 4i from part b in cartesian form? If so, how do I go about it (considering there are 3 solutions)?

Something to consider: Cube roots are a power of 1/3, so an exponential operation. This would mean that an exponential form of the number to be operated on would be easiest to work with, as the index laws can be easily applied. To do this we need to make use of Euler's formula: \displaystyle \begin{align*} e^{i \theta} = \cos{(\theta)} + i\sin{(\theta)} \end{align*}.

Now dealing with the number \displaystyle \begin{align*} -4\sqrt{3} - 4i = 4 \left( -\sqrt{3} - i \right) \end{align*}, if \displaystyle \begin{align*} z = -\sqrt{3} - i \end{align*}, then

\displaystyle \begin{align*} |z| &= \sqrt{ \left( -\sqrt{3} \right) ^2 + (-1)^2 } \\ &= \sqrt{3 + 1} \\ &= \sqrt{4} \\ &= 2 \end{align*}

and since z is in the third quadrant, we should find out what the equivalent angle is in the first quadrant and then convert it. The equivalent number in the first quadrant is \displaystyle \begin{align*} \sqrt{3} + i \end{align*}, so its angle is \displaystyle \begin{align*} \arctan{ \left( \frac{1}{\sqrt{3}} \right) } = \frac{\pi}{6} \end{align*}. This means that \displaystyle \begin{align*} \textrm{Arg}\,{(z)} = \frac{\pi}{6} - \pi = -\frac{5\pi}{6} \end{align*}. This is just the principal argument though, all possible arguments would be found by getting the equivalent angle by adding or subtracting multiples of \displaystyle \begin{align*} 2\pi \end{align*}. So \displaystyle \begin{align*} \arg{(z)} = 2\pi n - \frac{5\pi}{6} \end{align*}, where \displaystyle \begin{align*} n \in \mathbf{Z} \end{align*}.

Therefore \displaystyle \begin{align*} -4\sqrt{3} - 4i = 4 \left( -\sqrt{3} - i \right) = 4 \cdot 2 e^{i\left( 2\pi n - \frac{5\pi}{6} \right) } = 8e^{i\left( 2\pi n - \frac{5\pi}{6} \right) } \end{align*}.

So what do you think you'll get when you try to find its cube roots, i.e. \displaystyle \begin{align*} \left[ 8e^{i\left( 2\pi n - \frac{5\pi}{6} \right) } \right] ^{\frac{1}{3}} \end{align*}?
• Mar 3rd 2014, 02:42 PM
Fratricide
Re: Complex Numbers
Ok, so I've got the 3 roots. They are: 2cis(-5π/18), 2cis(7π/18) and 2cis(-17π/18). But how do I incorporate them into part (c) to solve for w?

(Sorry if I my original post was a bit ambiguous, but part (c) is the only part I need help with.)
• Mar 3rd 2014, 02:49 PM
Prove It
Re: Complex Numbers
Start by expanding \displaystyle \begin{align*} (z - w)^3 \end{align*} to give \displaystyle \begin{align*} z^3 - 3z^2\,w + 3z\,w^2 - w^3 \end{align*}, giving \displaystyle \begin{align*} z^3 - 3z^2w + 3z\,w^2 - w^3 = z^3 - 3\sqrt{3}i \, z^2 - 9z + 3\sqrt{3}i \end{align*}, giving

\displaystyle \begin{align*} -3w = -3\sqrt{3}i, 3w^2 = -9, -w^3 = 3\sqrt{3} \end{align*}

Can you get a consistent value for w?
• Mar 3rd 2014, 05:00 PM
Fratricide
Re: Complex Numbers
That'd be √3i, which would make my answer (z - √3i) = -4√3 - 4i.

Part (d) of the question reads, "Hence find in exact cartesian form the solutions of the equation z3 - 3√3iz2 - 9z + (3√3 + 4)i + 4√3 = 0".

Am I right in saying that the equation can be rewritten as (z - √3i)3 = -4√3 - 4i? Also, if we solve for z, would my answers be (using my answers from part (b)) 2cos(-5π/18) + (2sin(-5π/18) + √3)i, 2cos(7π/18) + (2sin(7π/18) + √3)i and 2cos(-17π/18) + (2sin(-17π/18) + √3)i?