Complex Numbers - writing in polar form

[srg263]

Hello everyone,
I have a complex number problem that i would greatly appreciate some help with. Thanks in advance to anyone offering their time to make a contribution.

Q) Write the following in polar form:
View attachment 6675I have attempted the question (please see my working below) and have been advised that i have missed a property of complex numbers that i should exploit to answer this problem, and also the argument is wrong.

View attachment 6676

Any assistance with corrections and the property of complex numbers that i should be using would be great as I'm quite lost.

Many thanks :-)

 

[MarkFL]

I think I would begin by writing:

\(\displaystyle 1+\sqrt{3}i=2\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=2\left(\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)\right)=2e^{\frac{\pi}{3}i}\)

And so:

\(\displaystyle \left(1+\sqrt{3}i\right)^8=2^8e^{\frac{8\pi}{3}i}\)

Then:

\(\displaystyle -1-i=\sqrt{2}\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\right)=\sqrt{2}\left(\cos\left(\frac{5\pi}{4}\right)+i\sin\left(\frac{5\pi}{4}\right)\right)=\sqrt{2}e^{\frac{5\pi}{4}i}\)

And so:

\(\displaystyle \left(-1-i\right)^5=2^{\frac{5}{2}}e^{\frac{25\pi}{4}i}\)

Thus:

\(\displaystyle \frac{\left(1+\sqrt{3}i\right)^8}{\left(-1-i\right)^5}=\frac{2^8e^{\frac{8\pi}{3}i}}{2^{\frac{5}{2}}e^{\frac{25\pi}{4}i}}=2^{\frac{11}{2}}e^{-\frac{43\pi}{12}i}=32\sqrt{2}e^{\frac{5\pi}{12}}\)

And so, in polar form, we have:

\(\displaystyle r=32\sqrt{2}\)

\(\displaystyle \theta=\frac{5\pi}{12}\)

I'm not sure what property of complex numbers is supposed to be used here, but that's how I'd work the problem. :D

 

[srg263]

Hi MarkFL,

Thanks so much for your response - really appreciated!

I've been looking further into the properties comment and found this in my text...

Properties of complex numbers:

\(\displaystyle \frac{\text{cis}(\theta)}{\text{cis}(\partial)} = \text{cis}(\theta-\partial)\)

and \(\displaystyle z=2\,\text{cis}\left(30^{\circ}\right)\)

\(\displaystyle z^3=2^3\text{cis}\left(3\cdot30^{\circ}\right)\)

I think my teacher was trying to show how you can manipulate division and powers with complex numbers.
I'm finding it tricky to understand though as the cis form of complex numbers very much confuse me.

 

[MarkFL]

Hi MarkFL,

Thanks so much for your response - really appreciated!

I've been looking further into the properties comment and found this in my text...

Properties of complex numbers:

\(\displaystyle \frac{\text{cis}(\theta)}{\text{cis}(\partial)} = \text{cis}(\theta-\partial)\)

and \(\displaystyle z=2\,\text{cis}\left(30^{\circ}\right)\)

\(\displaystyle z^3=2^3\text{cis}\left(3\cdot30^{\circ}\right)\)

I think my teacher was trying to show how you can manipulate division and powers with complex numbers.
I'm finding it tricky to understand though as the cis form of complex numbers very much confuse me.

I edited your post to wrap [MATH][/MATH] tags around your $\LaTeX$ code, and made a few other minor edits.

Well, I think those properties can be said to be a direct consequence of Euler's formula:

\(\displaystyle \text{cis}(\theta)=e^{\theta i}\)

And this is the formula you were using...you only made an error computing the argument, as you said. :D

 

[MarkFL]

I had an error in post #2, which I have now corrected.

 

[srg263]

Thank you! Apologies for you needing to correct the post. I can't figure out the symbol/command function with coding? I can see the symbol command set to click on on the right but when i click, the symbols don't go into this text box I'm writing in? So sorry but I'm really useless with IT, I've always preferred paper and pen!

 

[MarkFL]

Thank you! Apologies for you needing to correct the post. I can't figure out the symbol/command function with coding? I can see the symbol command set to click on on the right but when i click, the symbols don't go into this text box I'm writing in? So sorry but I'm really useless with IT, I've always preferred paper and pen!

No worries...using $\LaTeX$ takes a bit to get the hang of.

But, when you click a symbol/command in the Quick $\LaTeX$ element, it should appear in the post editor, as long as that's where your cursor is. If your cursor isn't active, then it should also appear in the post editor anyway (at least that's what it does for me, and is how we coded it). What OS/browser combination are you using?

 

[srg263]

No worries...using $\LaTeX$ takes a bit to get the hang of.

But, when you click a symbol/command in the Quick $\LaTeX$ element, it should appear in the post editor, as long as that's where your cursor is. If your cursor isn't active, then it should also appear in the post editor anyway (at least that's what it does for me, and is how we coded it). What OS/browser combination are you using?

I'm using Chrome and OS X Yosemite.

Am i correct in understand i need to add in coding around the symbol/commands? I am placing my cursor in the text editor prior to clicking the symbols. I've had a go below, hopefully that works!\(\displaystyle \infty \ne \alpha \)

 

[MarkFL]

I'm using Chrome and OS X Yosemite.

Am i correct in understand i need to add in coding around the symbol/commands? I am placing my cursor in the text editor prior to clicking the symbols. I've had a go below, hopefully that works!\(\displaystyle \infty \ne \alpha \)

Yes, the $\LaTeX$ code has to be wrapped with tags in order for it to be parsed correctly. :D

So, is the Quick $\LaTeX$ element working, and all you needed to do was wrap the code in the tags?

 

[srg263]

Following on from this question MarkFL, i had a final problem i was hoping you (or other users) may be able to assist with.

Express z4 = \(\displaystyle \sqrt{-3} + i \) in polar form and hence solve the equation z^2 = z4.

So far i have found that the polar form of z4 = \(\displaystyle 2 e^{}\frac{5i\pi}{6} \)

Following from this...

z^2 = z4
z^2 = \(\displaystyle 2 e^{}\frac{5i\pi}{6} \)
z^2 = \(\displaystyle \sqrt{2}² e^{\frac{5i\pi}{12}}² \)

But I'm somewhat stuck in where to go from here.
I'm of the thinking that two answers should arise, but I'm not sure how to work this out and also explain the reasoning behind why there should be two answers, which i need to do?

Many thanks for your time and help so far!

- - - Updated - - -

Yes, the $\LaTeX$ code has to be wrapped with tags in order for it to be parsed correctly. :D

So, is the Quick $\LaTeX$ element working, and all you needed to do was wrap the code in the tags?

Hi MarkFL,

Yes i think the $\LaTeX$ element is definitely working - it is my ineptitude that is the problem haha.

 

[MarkFL]

Following on from this question MarkFL, i had a final problem i was hoping you (or other users) may be able to assist with.

Express z4 = \(\displaystyle \sqrt{-3} + i \) in polar form and hence solve the equation z^2 = z4.

So far i have found that the polar form of z4 = \(\displaystyle 2 e^{}\frac{5i\pi}{6} \)

Following from this...

z^2 = z4
z^2 = \(\displaystyle 2 e^{}\frac{5i\pi}{6} \)
z^2 = \(\displaystyle \sqrt{2}² e^{\frac{5i\pi}{12}}² \)

But I'm somewhat stuck in where to go from here.
I'm of the thinking that two answers should arise, but I'm not sure how to work this out and also explain the reasoning behind why there should be two answers, which i need to do?

Many thanks for your time and help so far!

- - - Updated - - -
Hi MarkFL,

Yes i think the $\LaTeX$ element is definitely working - it is my ineptitude that is the problem haha.

You don't want to use the [SUP][/SUP] tags for exponents, you want to use the caret symbol instead.

Now, I am thinking the follow-up problem is:

Express \(\displaystyle z_4=\sqrt{3}+i\) in polar form, and hence solve the equation $z^2=z_4$

So, I would first write:

\(\displaystyle z_4=\sqrt{3}+i=2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=2\text{cis}\left(\frac{\pi}{3}\right)=2e^{\frac{\pi}{3}i}\)

But, notice we can also write:

\(\displaystyle z_4=\sqrt{3}+i=2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=2\text{cis}\left(\frac{7\pi}{3}\right)=2e^{\frac{7\pi}{3}i}\)

Because of the exponent of 2 on $z$ in the equation we are solving, we want to look at angles in the interval $0\le\theta<2\cdot2\pi$, because we we multiply the exponent on $e$ by the reciprocal of the exponent, we will bring all angles into the interval $0\le\dfrac{\theta}{2}<2\pi$.

And so we find:

\(\displaystyle z=\sqrt{2}e^{\frac{\pi}{6}i}\)

and:

\(\displaystyle z=\sqrt{2}e^{\frac{7\pi}{6}i}\)

Does that make sense?

 

[srg263]

You don't want to use the [SUP][/SUP] tags for exponents, you want to use the caret symbol instead.

Now, I am thinking the follow-up problem is:

Express \(\displaystyle z_4=\sqrt{3}+i\) in polar form, and hence solve the equation $z^2=z_4$

So, I would first write:

\(\displaystyle z_4=\sqrt{3}+i=2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=2\text{cis}\left(\frac{\pi}{3}\right)=2e^{\frac{\pi}{3}i}\)

But, notice we can also write:

\(\displaystyle z_4=\sqrt{3}+i=2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=2\text{cis}\left(\frac{7\pi}{3}\right)=2e^{\frac{7\pi}{3}i}\)

Because of the exponent of 2 on $z$ in the equation we are solving, we want to look at angles in the interval $0\le\theta<2\cdot2\pi$, because we we multiply the exponent on $e$ by the reciprocal of the exponent, we will bring all angles into the interval $0\le\dfrac{\theta}{2}<2\pi$.

And so we find:

\(\displaystyle z=\sqrt{2}e^{\frac{\pi}{6}i}\)

and:

\(\displaystyle z=\sqrt{2}e^{\frac{7\pi}{6}i}\)

Does that make sense?

Hi MARKFL,

Many thanks for that response. I'm still unfortunately a little confused.

Also, i noticed you used z4 = \(\displaystyle \sqrt{3} +i \) when the question is \(\displaystyle -\sqrt{3} + i \)
sorry if my question didn't make this clear with my poor formatting. As such, I'm thinking you're working would be slightly out as you didn't use \(\displaystyle -\sqrt{3} \).

I was going through my text and had the feeling perhaps this question has something to do with quadratic equations with complex coefficients? As in - if can find the square root of any complex number, the the quadratic formula can be used - and hence two answers?

Sorry for the ongoing questions, I've found complex numbers something I'm really struggling with.

 

[MarkFL]

Hi MARKFL,

Many thanks for that response. I'm still unfortunately a little confused.

Also, i noticed you used z4 = \(\displaystyle \sqrt{3} +i \) when the question is \(\displaystyle -\sqrt{3} + i \)

You had $\sqrt{-3}$ and I assumed you meant $\sqrt{3}$...I didn't realize you really meant $-\sqrt{3}$. Yes, this changes things...

So, I would first write:

\(\displaystyle z_4=-\sqrt{3}+i=2\left(\frac{-\sqrt{3}}{2}+\frac{1}{2}i\right)=2\text{cis}\left(\frac{5\pi}{6}\right)=2e^{\frac{5\pi}{6}i}\)

But, notice we can also write:

\(\displaystyle z_4=-\sqrt{3}+i=2\left(\frac{-\sqrt{3}}{2}+\frac{1}{2}i\right)=2\text{cis}\left(\frac{17\pi}{6}\right)=2e^{\frac{17\pi}{6}i}\)

Because of the exponent of 2 on $z$ in the equation we are solving, we want to look at angles in the interval $0\le\theta<2\cdot2\pi$, because we we multiply the exponent on $e$ by the reciprocal of the exponent, we will bring all angles into the interval $0\le\dfrac{\theta}{2}<2\pi$.

And so we find:

\(\displaystyle z=\sqrt{2}e^{\frac{5\pi}{12}i}\)

and:

\(\displaystyle z=\sqrt{2}e^{\frac{17\pi}{12}i}\)

 

[srg263]

Thank you for updating the solution, greatly appreciated.

If possible, i was wondering if you could explain how \(\displaystyle z4 = -\sqrt{3} + i \) can be written in two different ways as you have shown below. I don't understand this step. Many thanks.
View attachment 6678

 

[MarkFL]

Thank you for updating the solution, greatly appreciated.

If possible, i was wondering if you could explain how \(\displaystyle z4 = -\sqrt{3} + i \) can be written in two different ways as you have shown below. I don't understand this step. Many thanks.

Well, what I did was found $0\le \theta<2\pi$ such that:

\(\displaystyle \text{cis}(\theta)=\frac{-\sqrt{3}}{2}+\frac{1}{2}i\)

And then we use the periodicity of the cis() function:

\(\displaystyle \text{cis}(\theta+2k\pi)=\text{cis}(\theta)\) where \(\displaystyle k\in\mathbb{Z}\)

To find the solution in the next cycle ($k=1$), so that we have two arguments in $0\le\theta<4\pi$...because when we reduce the argument by dividing by 2 when we solve for $z$, we will have two arguments in $0\le\theta<2\pi$.

Does that make sense?

 

[srg263]

Well, what I did was found $0\le \theta<2\pi$ such that:

\(\displaystyle \text{cis}(\theta)=\frac{-\sqrt{3}}{2}+\frac{1}{2}i\)

And then we use the periodicity of the cis() function:

\(\displaystyle \text{cis}(\theta+2k\pi)=\text{cis}(\theta)\) where \(\displaystyle k\in\mathbb{Z}\)

To find the solution in the next cycle ($k=1$), so that we have two arguments in $0\le\theta<4\pi$...because when we reduce the argument by dividing by 2 when we solve for $z$, we will have two arguments in $0\le\theta<2\pi$.

Does that make sense?

Yes with that explanation and also revisiting the textbook this is starting to make sense now. Thanks so much, you've been so helpful!