# Thread: [SOLVED] Rectangular form of a complex number

1. ## [SOLVED] Rectangular form of a complex number

You know, I'm very glad I'm going through this little book. I thought I knew everything about doing these, but I had never considered a problem like this one.

The problem is to put
$\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}$
in rectangular form.

I'm kinda stuck on this one.

Before I mention what I've tried, let me say what the book's answer for this one is:
$(2 - \sqrt{3})^{12}$

My plan of attack was to write $\sqrt{3 - i}$ in rectangular form, giving me
$\sqrt{3 - i} = 10^{1/4}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) + i \cdot 10^{1/4}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right )$
(By the way, this approach doesn't use that there are actually two values for $\sqrt{3 - i}$. I'm not sure which one to pick, if it even makes a difference.)

Now that I've got that, then I can rewrite the original problem:
$\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24} =$ $\left [ 1 - \frac{10^{1/4}}{2}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) - i \cdot \frac{10^{1/4}}{2}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) \right ] ^{24}$

Now, believe it or not I can actually do something with this, but it's too complicated for me to bother with the LaTeX. Suffice it to say I can put the base in trigonometric form and use De Moivre's theorem to finish up the problem. But the answer is inordinately messy and I'm frankly not sure I got all of it right with all the picky details to keep track of.

And it certainly isn't the nice form $(2 - \sqrt{3})^{12}$.

Any thoughts as to how to approach this? Thanks!

-Dan

2. Originally Posted by topsquark
You know, I'm very glad I'm going through this little book. I thought I knew everything about doing these, but I had never considered a problem like this one.

The problem is to put
$\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}$
in rectangular form.

I'm kinda stuck on this one.

Before I mention what I've tried, let me say what the book's answer for this one is:
$(2 - \sqrt{3})^{12}$

My plan of attack was to write $\sqrt{3 - i}$ in rectangular form, giving me
$\sqrt{3 - i} = 10^{1/4}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) + i \cdot 10^{1/4}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right )$
(By the way, this approach doesn't use that there are actually two values for $\sqrt{3 - i}$. I'm not sure which one to pick, if it even makes a difference.)

Now that I've got that, then I can rewrite the original problem:
$\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24} =$ $\left [ 1 - \frac{10^{1/4}}{2}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) - i \cdot \frac{10^{1/4}}{2}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) \right ] ^{24}$

Now, believe it or not I can actually do something with this, but it's too complicated for me to bother with the LaTeX. Suffice it to say I can put the base in trigonometric form and use De Moivre's theorem to finish up the problem. But the answer is inordinately messy and I'm frankly not sure I got all of it right with all the picky details to keep track of.

And it certainly isn't the nice form $(2 - \sqrt{3})^{12}$.

Any thoughts as to how to approach this? Thanks!

-Dan
My first thought was to see what $\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{2}$ was - mind you, knowing the answer helped suggest that thought.

Imagine my disappointment when I got chicken vomit as the answer. Then I decided to put $\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}$ in my TI-89 ......

Imagine my surprise when I got an answer with a non-zero imaginary part .....

I'd suggest the question has a typo in it .....

3. Originally Posted by topsquark
You know, I'm very glad I'm going through this little book. I thought I knew everything about doing these, but I had never considered a problem like this one.

The problem is to put
$\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}$
in rectangular form.

I'm kinda stuck on this one.

Before I mention what I've tried, let me say what the book's answer for this one is:
$(2 - \sqrt{3})^{12}$

My plan of attack was to write $\sqrt{3 - i}$ in rectangular form, giving me
$\sqrt{3 - i} = 10^{1/4}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) + i \cdot 10^{1/4}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right )$
(By the way, this approach doesn't use that there are actually two values for $\sqrt{3 - i}$. I'm not sure which one to pick, if it even makes a difference.)

Now that I've got that, then I can rewrite the original problem:
$\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24} =$ $\left [ 1 - \frac{10^{1/4}}{2}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) - i \cdot \frac{10^{1/4}}{2}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) \right ] ^{24}$

Now, believe it or not I can actually do something with this, but it's too complicated for me to bother with the LaTeX. Suffice it to say I can put the base in trigonometric form and use De Moivre's theorem to finish up the problem. But the answer is inordinately messy and I'm frankly not sure I got all of it right with all the picky details to keep track of.

And it certainly isn't the nice form $(2 - \sqrt{3})^{12}$.

Any thoughts as to how to approach this? Thanks!

-Dan
A thought for checking the consistency of question and answer ...... a 24th root of the answer should be $1 - \frac{\sqrt{3 - i}}{2}$, right?

4. Originally Posted by topsquark
You know, I'm very glad I'm going through this little book. I thought I knew everything about doing these, but I had never considered a problem like this one.

The problem is to put
$\left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}$
in rectangular form.

I'm kinda stuck on this one.

Before I mention what I've tried, let me say what the book's answer for this one is:
$(2 - \sqrt{3})^{12}$
$\left | 1 - \frac{\sqrt{3 - i}}{2} \right |\approx 0.187755$

$0.187755^{24} \approx 3.68317 \times 10^{-18}$

$2 - \sqrt{3} \approx 0.267949$

$0.267949^{12} \approx 1.36971\times 10^{-7}$

So I agree with mr fantastic, you have a typo.

RonL

5. Originally Posted by CaptainBlack
$\left | 1 - \frac{\sqrt{3 - i}}{2} \right |\approx 0.187755$

$0.187755^{24} \approx 3.68317 \times 10^{-18}$

$2 - \sqrt{3} \approx 0.267949$

$0.267949^{12} \approx 1.36971\times 10^{-7}$

So I agree with mr fantastic, you have a typo.

RonL
Nice back-of-the-envelope confirmation, CaptainBlack. So the question now is:

What's the typo, that is, what was the intended question? Working back from the answer would probably get it .....

6. Originally Posted by mr fantastic
So the question now is:

What's the typo, that is, what was the intended question?
As a first guess, $\left ( 1 - \frac{\sqrt3 - i}{2} \right ) ^{24}$. That has the correct absolute value, at any rate.

7. Originally Posted by Opalg
As a first guess, $\left ( 1 - \frac{\sqrt3 - i}{2} \right ) ^{24}$. That has the correct absolute value, at any rate.
Great get, Opalq ol' bean! That is indeed the answer to what question gives the answer.

8. Originally Posted by Opalg
As a first guess, $\left ( 1 - \frac{\sqrt3 - i}{2} \right ) ^{24}$. That has the correct absolute value, at any rate.
There we go! And that makes the problem MUCH simpler. (I'll double check the problem, but I think I typed it in here right.) The trouble with my seeing that on my own is that I was doing problems like $\sqrt[8]{3 - 4i}$ as well, so it could very well have been the problem as I wrote it. But I can do the revised problem without any undue difficulty.

-Dan

9. Thank you all for your help!

-Dan