ArGand Diagrams

[Aleena23]

Can u do question number 11? It is attached below? I am just a bit puzzled if I use root 3 or minus root3?
whohc one should I use?

My working out….

- w+z =( k-4 )+ (5k+1)i

- 2pi/3= 120 degrees so it is on the left top quadrant.

- argument = 2pi/3. so the right angled triangle will have an angle
60 degrees = pi/3

- since we r in the top left quadrant the adjacent will need to be negative so it will become 4-k.

- soo tan(pi/3) = Opposite / Adjacent.

- soo root3 = 5k+1 / 4-k
= 4root3 - k root 3 = 5k+1
5k+k root 3 = 4 root 3 - 1
k (5 + root 3) = 4 root 3 - 1
k =( 4 root 3 - 1) / (5+ root 3)

Therefore k = (-17 + 21 root 3) / (22)

 

[Dr.Peterson]

Can u do question number 11? It is attached below? I am just a bit puzzled if I use root 3 or minus root3?
whohc one should I use?

My working out….

- w+z =( k-4 )+ (5k+1)i

- 2pi/3= 120 degrees so it is on the left top quadrant.

- argument = 2pi/3. so the right angled triangle will have an angle
60 degrees = pi/3

- since we r in the top left quadrant the adjacent will need to be negative so it will become 4-k.

- soo tan(pi/3) = Opposite / Adjacent.

- soo root3 = 5k+1 / 4-k
= 4root3 - k root 3 = 5k+1
5k+k root 3 = 4 root 3 - 1
k (5 + root 3) = 4 root 3 - 1
k =( 4 root 3 - 1) / (5+ root 3)

Therefore k = (-17 + 21 root 3) / (22)

Your work is correct, though it could be written a little better (parentheses in "5k+1 / 4-k", for example).

You correctly use sqrt(3), because that is tan(pi/3). At that point you are using the reference angle, which is in the first quadrant.

You could have directly used tan(2pi/3), which would have been -sqrt(3), being in the second quadrant.

Try plotting w, z, and w+z to see that you have a valid solution.

 

[Aleena23]

View attachment 31642

Your work is correct, though it could be written a little better (parentheses in "5k+1 / 4-k", for example).

You correctly use sqrt(3), because that is tan(pi/3). At that point you are using the reference angle, which is in the first quadrant.

You could have directly used tan(2pi/3), which would have been -sqrt(3), being in the second quadrant.

Try plotting w, z, and w+z to see that you have a valid solution.

Hi sir so I have to use tan(3pi/3)? Nit Not tan(pi/3)?

 

[Dr.Peterson]

Hi sir so I have to use tan(3pi/3)? Not tan(pi/3)?

Why do you say that? Please explain your thinking.

 

[Aleena23]

Why do you say that? Please explain your thinking.

As in the angle isn’t being negated, only the adjacent is. And we r using tan so there needs to be a right angled triangle involved and by using (tan 2pi/3) which means tan(2*180/3) which equal 120 degrees. In a right angled triankge there always need to be angles between 0 and 90. Unless we r working with sine rule and cosine rule,

we need to use tan(pi/3) as that will make a right angled triangle. And that is the angle alpha while theta is the argument (2pi/3). I have an image below sir.

 

[Dr.Peterson]

As in the angle isn’t being negated, only the adjacent is. And we r using tan so there needs to be a right angled triangle involved and by using (tan 2pi/3) which means tan(2*180/3) which equal 120 degrees. In a right angled triankge there always need to be angles between 0 and 90. Unless we r working with sine rule and cosine rule,

we need to use tan(pi/3) as that will make a right angled triangle. And that is the angle alpha while theta is the argument (2pi/3). I have an image below sir.

Yes, what you marked as [imath]\alpha[/imath] is the reference angle, and it equal to [imath]\frac{\pi}{3}[/imath].

So I ask again: Why do you say, "so I have to use tan(3pi/3)? Not tan(pi/3)?" Nothing you say suggests that.

 

[Aleena23]

Yes, what you marked as [imath]\alpha[/imath] is the reference angle, and it equal to [imath]\frac{\pi}{3}[/imath].

So I ask again: Why do you say, "so I have to use tan(3pi/3)? Not tan(pi/3)?" Nothing you say suggests that.

1) I meant to ask …“so I have to use *tan(2pi/3)*? Am I not supposed it be using Tan(pi/3) ?
cos we r working with right angle triangle right?

2) sir why do I have to use tan(2pi/3)? why can’t I use tan(pi/3)?

3) is it wrong if I use tan(pi/3)?

 

[Dr.Peterson]

1) I meant to ask …“so I have to use *tan(2pi/3)*? Am I not supposed it be using Tan(pi/3) ?
cos we r working with right angle triangle right?

2) sir why do I have to use tan(2pi/3)? why can’t I use tan(pi/3)?

3) is it wrong if I use tan(pi/3)?

What did I say before, about your work using tan(pi/3)? Here it is:

Your work is correct, ...

You correctly use sqrt(3), because that is tan(pi/3). At that point you are using the reference angle, which is in the first quadrant.

You could have directly used tan(2pi/3), which would have been -sqrt(3), being in the second quadrant.

Try plotting w, z, and w+z to see that you have a valid solution.

So I never said either way is wrong!!

To be as clear as I can be:

The problem is,


There are (at least) two ways to answer this.

First, [imath]w+z=(k+i)+(-4+5ki) = (k-4)+(5k+1)i[/imath].

Method 1 (mine, using general angles): Since the argument is [imath]\frac{2\pi}{3}[/imath], an angle in the second quadrant, its tangent is [imath]-\sqrt{3}[/imath], which has to be equal to [imath]\frac{5k+1}{k-4}[/imath]. So we need to solve [imath]\frac{5k+1}{k-4}=-\sqrt{3}[/imath]. Therefore [math]5k+1=-\sqrt{3}(k-4)[/math]
Method 2 (yours, as I understand it, using a right triangle): Since the argument is [imath]\frac{2\pi}{3}[/imath], an angle in the second quadrant, the reference angle is [imath]\frac{\pi}{3}[/imath], and in the reference triangle in the first quadrant, the horizontal leg is the negative of that in the second quadrant, so the ratio [imath]\frac{5k+1}{-(k-4)}[/imath] must equal the tangent of the reference angle, namely [imath]\sqrt{3}[/imath]. So we need to solve [imath]\frac{5k+1}{4-k}=\sqrt{3}[/imath]. Therefore [math]5k+1=\sqrt{3}(4-k)[/math]
Continuing from either method, [math](5+\sqrt{3})k=-1+4\sqrt{3}\implies k=\frac{-1+4\sqrt{3}}{5+\sqrt{3}}=\frac{-17+21\sqrt{3}}{22}[/math]
So you have to use the given angle, but can use the reference angle or not. It makes no difference.

So, what are the answers to your questions?

By the way, I have a couple comments about your writing. First, in the original post, you have "-" at the start of many lines, which can be easily read as negative signs. That is confusing. Second, in this response you used abbreviations, "cos we r working", which are particularly hazardous when you're talking about trigonometry! Our guidelines recommend avoiding this, for good reasons!

 

[Aleena23]

What did I say before, about your work using tan(pi/3)? Here it is:

So I never said either way is wrong!!

To be as clear as I can be:

The problem is,
View attachment 31675

There are (at least) two ways to answer this.

First, [imath]w+z=(k+i)+(-4+5ki) = (k-4)+(5k+1)i[/imath].

Method 1 (mine, using general angles): Since the argument is [imath]\frac{2\pi}{3}[/imath], an angle in the second quadrant, its tangent is [imath]-\sqrt{3}[/imath], which has to be equal to [imath]\frac{5k+1}{k-4}[/imath]. So we need to solve [imath]\frac{5k+1}{k-4}=-\sqrt{3}[/imath]. Therefore [math]5k+1=-\sqrt{3}(k-4)[/math]
Method 2 (yours, as I understand it, using a right triangle): Since the argument is [imath]\frac{2\pi}{3}[/imath], an angle in the second quadrant, the reference angle is [imath]\frac{\pi}{3}[/imath], and in the reference triangle in the first quadrant, the horizontal leg is the negative of that in the second quadrant, so the ratio [imath]\frac{5k+1}{-(k-4)}[/imath] must equal the tangent of the reference angle, namely [imath]\sqrt{3}[/imath]. So we need to solve [imath]\frac{5k+1}{4-k}=\sqrt{3}[/imath]. Therefore [math]5k+1=\sqrt{3}(4-k)[/math]
Continuing from either method, [math](5+\sqrt{3})k=-1+4\sqrt{3}\implies k=\frac{-1+4\sqrt{3}}{5+\sqrt{3}}=\frac{-17+21\sqrt{3}}{22}[/math]
So you have to use the given angle, but can use the reference angle or not. It makes no difference.

So, what are the answers to your questions?

By the way, I have a couple comments about your writing. First, in the original post, you have "-" at the start of many lines, which can be easily read as negative signs. That is confusing. Second, in this response you used abbreviations, "cos we r working", which are particularly hazardous when you're talking about trigonometry! Our guidelines recommend avoiding this, for good reasons!

The answer is That I can use both but differently. Thank you sir so much! And I won’t use abbreviations and no more minus looking signs! Sorry about that!

 

[pka]

Can u do question number 11? It is attached below? I am just a bit puzzled if I use root 3 or minus root3?
whohc one should I use?

First, I want to second every thing Prof, Peterson said about your hand-writing as well as your use of English. This is not some trivial chat-site. So please learn to post readable text.
If [imath]z=a+b\bf i[/imath] here [imath]a<0~\&~b>0[/imath] then [imath]z\in II[/imath] or quadrant two.
As such the argument is [imath]\arg(z)=\pi -\arctan\left(\left|\dfrac{b}{a}\right|\right)[/imath]
From what you were given in this question, we know that [imath]w+z=(k-4)+(5k+1)\bf i[/imath].
We also know that [imath]\arg(w+z)\in II[/imath] So we get [imath]\pi - \arctan \left( {\left| {\dfrac{{5k + 1}}{{k - 4}}} \right|} \right) = \dfrac{{2\pi }}{3}[/imath]
SEE HERE for the evaluation.
[imath][/imath][imath][/imath]

 

[Aleena23]

First, I want to second every thing Prof, Peterson said about your hand-writing as well as your use of English. This is not some trivial chat-site. So please learn to post readable text.
If [imath]z=a+b\bf i[/imath] here [imath]a<0~\&~b>0[/imath] then [imath]z\in II[/imath] or quadrant two.
As such the argument is [imath]\arg(z)=\pi -\arctan\left(\left|\dfrac{b}{a}\right|\right)[/imath]
From what you were given in this question, we know that [imath]w+z=(k-4)+(5k+1)\bf i[/imath].
We also know that [imath]\arg(w+z)\in II[/imath] So we get [imath]\pi - \arctan \left( {\left| {\dfrac{{5k + 1}}{{k - 4}}} \right|} \right) = \dfrac{{2\pi }}{3}[/imath]
SEE HERE for the evaluation.
[imath][/imath][imath][/imath]

Thank you sir /madam