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Complex Numbers

FatRat

I'm doing the leaving cert in 2014 and I'm finding Complex numbers to be very very difficult. In comparison to the rest of paper, I think it's ridiculously complicated. At the moment I'm stuck on this question with absolutely no clue on how to do it:

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ZorbaTehZ

Regarding what can be inferred from the three equations:

Assuming you know what polar form is, write [latex]z_1=e^{i \theta_1}[/latex] and [latex] z_2=e^{i \theta_2}[/latex]. Since [latex] i = e^{i \pi /2}[/latex] then rewrite [latex]z_2=iz_1[/latex] as an equality (up to addition of 2 pi) of the angles between [latex] \theta_1,\theta_2[/latex]. This will put (severe) restrictions on the possible choices.

For the second one, define a function [latex] f(t)=tz_1[/latex], then [latex]f[/latex] is a line through the origin and [latex] z_1[/latex] (provided [latex]z_1[/latex] is non-zero - in-fact [latex]z_1[/latex] has to be non-zero from the diagram). So if [latex]z_2= tz_1[/latex] then, what can be inferred about the possible locations of [latex]z_2[/latex]?

As for the final one, hint: parallelogram law.

Iderown

I would look at the second relationship first. It says that z3 and z1 are in the same "direction" from the origin. The real (imaginary) component of z3 is a scalar multiple of the real (imaginary) component of z1. The multiple, k, may be less than 1.

Then relationship 1. Remember that multiplying a complex number by i has the effect of rotating it's vector anti-clockwise by pi/2. Can you identify such a pair of vectors?

Now you know three of the four z. Really now you know all.

Relationship 3 can be used to verify your result.

(I'm going to have to learn how do make nice mathematical symbols!!)

764dak

Iderown's answer is better. The diagram uses the same scale for the axes. Z1 and Z2 would have the same length which (a^2 +b^2)^(1/2). Z2 is a rotation of Z1 by pi/2 or 90 degrees into the second quadrant. Z3 also has the same direction as Z2. You can deduce that the farthest point to the right is Z1, the second farthest is Z3 and the one farthest to the left Z2.

You can guess the value of k.

ZorbaTehZ

It is worth pointing out that the charter for this forum states that posters should give hints or guidance to questions, not answers. But 764dak's answer just gives it all away... oh well.

EDIT: I've just noticed a (rather minor) mistake in my post above (which I can't seem to edit for some reason): of course I should have written [latex]z_1=r_1 e^{i \theta_1}[/latex] and [latex] z_2=r_2e^{i \theta_2}[/latex] instead of [latex]z_1=e^{i \theta_1}[/latex] and [latex] z_2=e^{i \theta_2}[/latex]

Michael Collins

ZorbaTehZ wrote: »

It is worth pointing out that the charter for this forum states that posters should give hints or guidance to questions, not answers. But 764dak's answer just gives it all away... oh well.

EDIT: I've just noticed a (rather minor) mistake in my post above (which I can't seem to edit for some reason): of course I should have written [latex]z_1=r_1 e^{i \theta_1}[/latex] and [latex] z_2=r_2e^{i \theta_2}[/latex] instead of [latex]z_1=e^{i \theta_1}[/latex] and [latex] z_2=e^{i \theta_2}[/latex]

I don't think complex exponentials are on the Leaving Cert course though. Maybe restate in terms of de Moivre's identity:

[latex]z_1=r_1 e^{i \theta_1} = r_1 \left[ \cos(\theta_1) + i \sin(\theta_1)\right][/latex]

FatRat

I'm sorry I hadn't replied til now! I had figured it out after the first two post. I came up with exactly what 764dak said so thanks to the first two replies for that.

Michael Collins wrote: »

I don't think complex exponentials are on the Leaving Cert course though. Maybe restate in terms of de Moivre's identity:

[latex]z_1=r_1 e^{i \theta_1} = r_1 \left[ \cos(\theta) + i \sin(\theta)\right][/latex]

It isn't so this is where I got confused. I started learning about complex exponentials after online only to realise it isn't on the course

That said it still helps me somewhat. As for De Moivre's theorom, I still dont have that nailed down and that was going to be my next question! I understand the fundamentals of it but I need help mainly with the unit circle and using it to determine angles in pi form. I'll use this question as an example, it's the one that comes before the question I posted above:

I understand the first bit. Get the radius and write in basic polar form. But then when it comes to De moivre, I can never understand how one would get the angle in pi form. I know it has something to do with the different quadrants in the unit circle but it's all very vague to me. Never got my head around it!

Michael Collins

What did you get for part (i)?

FatRat

Michael Collins wrote: »

What did you get for part (i)?

r =[Square root of (Square root of 3)^2 + (-1)^2)] = Square root of 4 (I dont know how to do maths annotation, sorry!) = 2
^This is right, isn't it? I may be wrong of course.

So now I need to find the angle in radians. It's in the 2nd quadrant, isn't it? And you draw a triangle based on the x-axis up to the point w? I have the hypotenuse which is r, which = 2. I also have the opposite which is 1 (same thing as -1??) So I can use sin=O/H to get the angle of said triangle. Using that I get the angle as 50 degrees. Since it's in the 2nd quadrant, it's turning 180 degrees. So the angle I'm looking for is 180-50=130. And I need this in radians... no idea how I get that! All I know is Pi=180 degrees.

So in degrees (i) = 2(cos130 + isin130) .. BTW... I'm fairly sure all that I said above is wrong haha

764dak

http://www.mathwarehouse.com/trigonometry/radians/convert-degee-to-radians.php

This link shows you how to convert from degrees to radians and vice versa.

Michael Collins

FatRat wrote: »

r =[Square root of (Square root of 3)^2 + (-1)^2)] = Square root of 4 (I dont know how to do maths annotation, sorry!) = 2
^This is right, isn't it? I may be wrong of course.

We can all be wrong, but this is correct!

FatRat wrote: »

So now I need to find the angle in radians. It's in the 2nd quadrant, isn't it? And you draw a triangle based on the x-axis up to the point w? I have the hypotenuse which is r, which = 2. I also have the opposite which is 1 (same thing as -1??)

So far, so good. Yes, you're interested in the length, so can ignore the minus sign, which just indicates the direction.

FatRat wrote: »

So I can use sin=O/H to get the angle of said triangle. Using that I get the angle as 50 degrees. Since it's in the 2nd quadrant, it's turning 180 degrees. So the angle I'm looking for is 180-50=130. And I need this in radians... no idea how I get that! All I know is Pi=180 degrees.

You have the right idea I think, you just made a small error in the first angle. It should be

[latex]\displaystyle \sin(\phi) = \frac{\hbox{opp}}{\hbox{hyo}} = \frac{\sqrt{3}}{2}[/latex]

[latex]\displaystyle \phi = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60 ^{\circ} [/latex]

[latex]\theta = 180^{\circ} - 60^{\circ} = 120^{\circ}[/latex]

FatRat wrote: »

So in degrees (i) = 2(cos130 + isin130) .. BTW... I'm fairly sure all that I said above is wrong haha

Not all wrong at all, in fact almost all correct. The final answer for part (i) should be:

[latex] w = \displaystyle 2(\cos120^{\circ} + \hbox{i}\sin120^{\circ}) [/latex]

Now for the next part, you're told that

[latex] z^2 = \displaystyle -1 + \sqrt{3} \hbox{i} [/latex]

so clearly, this is the same as:

[latex] z^2 = \displaystyle 2(\cos120^{\circ} + \hbox{i}\sin120^{\circ}) [/latex].

Now try and apply de Moivre's identity. Or if you're not sure how, state it here as best you can.

FatRat

Michael Collins wrote: »

We can all be wrong, but this is correct!

So far, so good. Yes, you're interested in the length, so can ignore the minus sign, which just indicates the direction.

You have the right idea I think, you just made a small error in the first angle. It should be

[latex]\displaystyle \sin(\phi) = \frac{\hbox{opp}}{\hbox{hyo}} = \frac{\sqrt{3}}{2}[/latex]

[latex]\displaystyle \phi = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60 ^{\circ} [/latex]

[latex]\theta = 180^{\circ} - 60^{\circ} = 120^{\circ}[/latex]

Not all wrong at all, in fact almost all correct. The final answer for part (i) should be:

[latex] w = \displaystyle 2(\cos120^{\circ} + \hbox{i}\sin120^{\circ}) [/latex]

Now for the next part, you're told that

[latex] z^2 = \displaystyle -1 + \sqrt{3} \hbox{i} [/latex]

so clearly, this is the same as:

[latex] z^2 = \displaystyle 2(\cos120^{\circ} + \hbox{i}\sin120^{\circ}) [/latex].

Now try and apply de Moivre's identity. Or if you're not sure how, state it here as best you can.

I get all of that but I don't get why you're using square root of 3 as the opposite? I must be imagining the triangle drawn in a different way to you. Also when it comes to the angle... everytime I see questions like this done, the answer is never in degree format! Always in radians? And you use the unit circle or something weird like that the find it?

Michael Collins

FatRat wrote: »

I get all of that but I don't get why you're using square root of 3 as the opposite? I must be imagining the triangle drawn in a different way to you.

Yep, you seem to be!

It works like this:





        w = -1 + sqrt{3}
        |\
        | \
sqrt{3} |  \ 2
        |   \
        |   &#952;\
        &#175;&#175;&#175;&#175;&#175;&#175;
           1

So the -1 bit means you move to the left 1 unit, and the sqrt{3} bit means you move up, by sqrt{3} units.

So now, clearly, sin θ = opp/hyp = sqrt{3}/2

FatRat wrote: »

Also when it comes to the angle... everytime I see questions like this done, the answer is never in degree format! Always in radians? And you use the unit circle or something weird like that the find it?

By definition, there are [latex] 360 ^{\circ} [/latex] in a circle. This can also be expressed in radians, which is defined as the length of the circumference, divided by the radius. Cleary any full circle has circumference

[latex] \hbox{circumference length} = 2\pi r[/latex]

so then it must have

[latex] \displaystyle \hbox{angle in radians} = \frac{2\pi r}{r} = 2\pi.[/latex]

So there are [latex] 2\pi [/latex] radians in a full circle, hence

[latex] \displaystyle 360 \hbox{ degrees} = 2\pi \hbox{ radians}[/latex]

[latex] \displaystyle \Rightarrow 1 \hbox{ degree} = \frac{2\pi}{360} = \frac{\pi}{180} \hbox{ radians}[/latex]

So in the question above:

[latex] \displaystyle 120 \hbox{ degrees} = 120\left(\frac{\pi}{180}\right) = \frac{2\pi}{3}\hbox{ radians}[/latex]

Michael Collins

Actually – just realised I made a mistake in my last post. It should be inverse sin(sqrt(3)/2). Since the hypotenuse is 2 units long. This gives the angle of 60 degrees.

I've edited the post above now and corrected this. Sorry!

FatRat

Great! Think I have a good grasp of. Thanks everyone! Just wondering if there's anything else important in terms of complex numbers that I should watch out for? For example, ye said that mutiplying a number by i means it rotates it 90 degrees anti-clockwise. It's stuff like that I miss.

Iderown

FatRat wrote: »

Just wondering if there's anything else important in terms of complex numbers that I should watch out for? For example, ye said that mutiplying a number by i means it rotates it 90 degrees anti-clockwise. It's stuff like that I miss.

What do you think that DIVIDING a complex number by i does to its Argand diagram representation?

Here's a clue. I'm going to have to learn how to type nice mathematics here.

z/i = iz/(i squared) = iz/(-1) = -iz

ZorbaTehZ

FatRat wrote: »

Great! Think I have a good grasp of. Thanks everyone! Just wondering if there's anything else important in terms of complex numbers that I should watch out for? For example, ye said that multiplying a number by i means it rotates it 90 degrees anti-clockwise. It's stuff like that I miss.

Just do many questions and eventually you will start to see these things more easily.

Alternatively there is a book by Tristan Needham called Visual Complex Analysis that focuses on these diagrammatic ways of looking at complex equations.

gammy_knees

FatRat
You'll also need to know what multiplication and division of a+bi by c+di does, not just multiplication by i.
Go to projectmaths.ie and click on the student CD. In the LC section go to Strand 3-Numbers and scroll to the complex number section. Make sure you have Geogebra installed to run the files given. You'll find a lot there on how multiplication and division of C works.
Best of luck.

ray giraffe

ZorbaTehZ wrote: »

Alternatively there is a book by Tristan Needham called Visual Complex Analysis that focuses on these diagrammatic ways of looking at complex equations.

This book is not remotely appropriate for Leaving Cert students.

It's suitable for a minimum of second year undergraduates in a maths degree.

ZorbaTehZ

ray giraffe wrote: »

This book is not remotely appropriate for Leaving Cert students.

That is nonsense imo, but I suppose these things are a matter of taste.

TheBody

ray giraffe wrote: »

This book is not remotely appropriate for Leaving Cert students.

It's suitable for a minimum of second year undergraduates in a maths degree.

I agree. I own a copy. It's a beautiful book but too much for a Leaving Cert student.

bnt

Also, for a quick play with examples, try an octave online interpreter. The Octave (MATLAB) language is pretty easy. Quick complex example:

z = 4 + 3i
length = abs(z)
z*i
z/i

This defines a complex vector, finds its length, and then tries multiplication and division by i.