Help with These 5 complex number questions (1 Viewer)

MAK03

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Hey guys I just need some hep with a few complex number. I have named them from 1-5 so whichever one you can do please send the working out with the number of the questions
Thx in Advanced
 

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jimmysmith560

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For the question in the attachment "3.jpg", there is a similar version that is as follows:

1641190412732.png

The working to the above question is as follows:

1641190075125.png
1641190179720.png
1641190229423.png
1641190527075.png
1641190604013.png
1641190647214.png

A minor/simple step is missing at the end, which just involves multiplying the equation by 4 to end up with the equation that is specified in the question.

I hope this helps! :D
 

5uckerberg

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With the fourth image, the question you gonna ask yourself is how can we get from to . Now the first step we need to do is to recognise the double angle formula for the tangent.

Noting that the double angle formula for tangent is
. This the vital piece of the puzzle because we will be using it twice in this case.

For the question replace all the passive characters (betas) with the proactive characters (alphas) because as extension II students we have to be able to stand up to difficulty and overcome it

In doing so



Note one can say that

Next step



There you will obtain

Part b

The question you have to ask yourself is which number will form the hypotenuse with the side lengths of 7 and 24 and also note that in the question on its own means a restriction of . But remember what we did in part a we went into the realm of so therefore our restriction instead should be



Remember what you have here is to the fourth power which gives the length of 25 so thus your r should be . The angles is a whole different story. What we have to do is find and for 4 cycles okay so we have to add on that so therefore what we will obtain is

and

and


As specified the values of are found here





part c from what I see is calculator bashing to find the four roots of z. Not sure if calculators existed in the 1988 HSC for 4 units because I am curious.[/TEX][/TEX]
 
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5uckerberg

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The fifth question is simply the use of the conjugate root theorem because a complex number and its conjugate give us double the real portion of the complex number and that the product gives us the squared form of the real portion added to the squared form of the imaginary portion.
 
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Lith_30

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For image 1:

part i)
you just sub in x = 1 and into the equation and show that it equals to zero.

part ii)


part iii)
Not too sure about this, but by using the identity we can say that the diagonals of the shape formed by the roots are perpendicular and equal in length, which only a square can have.

part iv)
We already know that one root is 1 and by inspection we can see that .

we can sub beta into

we can then add this expression onto


Therefore the other two roots of the equation are and
 

5uckerberg

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The first question in part i is effectively asking whether and 1 are the roots of the equation. FIrst and foremost sub in . I guarantee it will give zero as an answer.

part ii is simply introducing the concept of sum and product of roots. As noticed here sum of roots for a polynomial is simply and product of roots for a polynomial of degree 4 is for any given polynomial of the form .

As noted the sum of the roots gives which if noted clearly b is the value
and using what we know , thus .

The product of 4 roots gives us . We know the product of so therefore the product should give us . Therefore, .

If you started with the topic proof this would be second nature for you because you are going to be using a building block from inequalities which is
. Thus, for this question this idea becomes .

There, .

My next question to you is how do you find .
Or should I say how do you find the square root of a negative number
 
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MAK03

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Thx sooooo much you basically answered all the questions
you guys are legends
 

5uckerberg

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For question 2 we are given

The modulus requires us to find . Now notice that you need to use the double angle formula to notice that . Multiply by 2 and you will have . Now 2 is just the square root of 4 right, so therefor multiply the RHS your modulus is just .

Since we are also finding the argument one important note is that for a complex number to be written in the modulus-argument form it will be written in the form of . Remember we have
as our modulus.

The argument will come from the structure of the complex number.

See here
Well in that case you can clearly see that
.
Using the double angle formula

then it is obvious that

and
now if we are finding the mod arg form then it becomes

Thus your final answer will be that the argument is

This part simply asks you to do this convert into mod arg form which we have found and then apply De Moivre's Theorem. Easy as cake.

part ii starts off asking us to use the real part of the complex number and combining De Moivre's Theorem and Binomial Theorem we will have
. As noticed for the modulus argument form we will also take the real part of the complex number giving us . Now as noted the modulus part when you raise to the power of 4 it gives us and for the argument part we use De Moivre's Theorem which will give us . Once you have that you can clearly see that from the question
giving us .
 
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5uckerberg

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For the sine version binomial theorem aggain but we are focusing on the imaginary side where you have the modulus times the imaginary part of the argument.
 

MAK03

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For question 2 we are given

The modulus requires us to find . Now notice that you need to use the double angle formula to notice that . Multiply by 2 and you will have . Now 2 is just the square root of 4 right, so therefor multiply the RHS your modulus is just .

Since we are also finding the argument one important note is that for a complex number to be written in the modulus-argument form it will be written in the form of . Remember we have
as our modulus.

The argument will come from the structure of the complex number.

See here
Well in that case you can clearly see that
.
Using the double angle formula

then it is obvious that

and
now if we are finding the mod arg form then it becomes

Thus your final answer will be that the argument is

This part simply asks you to do this convert into mod arg form which we have found and then apply De Moivre's Theorem. Easy as cake.

part ii starts off asking us to use the real part of the complex number and combining De Moivre's Theorem and Binomial Theorem we will have
. As noticed for the modulus argument form we will also take the real part of the complex number giving us . Now as noted the modulus part when you raise to the power of 4 it gives us and for the argument part we use De Moivre's Theorem which will give us . Once you have that you can clearly see that from the question
giving us .

Hey, sorry to be annoying but you re-explain how you obtained the expression on the LHS for ii
 

5uckerberg

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Hey, sorry to be annoying but you re-explain how you obtained the expression on the LHS for ii
Since you asked for it let's dive into this bad boy. From what we know we are going to start with
.

Simplifying we will have


Expanding will obtain


Collecting all the terms we are then left with


Now you might be going "huh, what is that supposed to mean we are going further from the goal"

I understand what you are feeling but now we can have some fun and destroy this godzilla looking problem.

Remembering the fact that we can now play with this problem

is simply


Expanding that we will have



Part 2 will come
 
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5uckerberg

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Which simplified itself will be


Simplifying further gives us


You can do the rest from here.

The other case is simply to do the same plethora from the first post but this time working with the imaginary terms and turning the cosines into sines and then using the fact that
 

5uckerberg

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What we should have is from the binomial expansion

Expanding it becomes





Notice when I said

It will become

Now do you recall this

Using that, the whole expression written in terms of z is just



Simplifying this will give us

 

5uckerberg

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Which can be factorised as




This is just

Further simplification will give us



Later on, it becomes
 

MAK03

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Sorry, I just saw this, thx soooooo much , you have been really helpfull
 

MAK03

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With the fourth image, the question you gonna ask yourself is how can we get from to . Now the first step we need to do is to recognise the double angle formula for the tangent.

Noting that the double angle formula for tangent is
. This the vital piece of the puzzle because we will be using it twice in this case.

For the question replace all the passive characters (betas) with the proactive characters (alphas) because as extension II students we have to be able to stand up to difficulty and overcome it

In doing so



Note one can say that

Next step



There you will obtain

Part b

The question you have to ask yourself is which number will form the hypotenuse with the side lengths of 7 and 24 and also note that in the question on its own means a restriction of . But remember what we did in part a we went into the realm of so therefore our restriction instead should be



Remember what you have here is to the fourth power which gives the length of 25 so thus your r should be . The angles is a whole different story. What we have to do is find and for 4 cycles okay so we have to add on that so therefore what we will obtain is

and

and


As specified the values of are found here





part c from what I see is calculator bashing to find the four roots of z. Not sure if calculators existed in the 1988 HSC for 4 units because I am curious.[/TEX][/TEX]
Hey again, could you please re-explain part b and c
 

5uckerberg

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Hey again, could you please re-explain part b and c
So for part b start with the fact that the length of is 25. and then the r which is length is the fourth root of 25 equalling . The argument is very simple as well. First we see that for is where so therefore, your four values are the last statement is simply just because by taking away you are back to where you started.

Why is this working so well, remember from the question it specified that "Given that and that one has to express 7+24i in the form then has to be which is simply as such and there the boundaries suddenly became so therefore, what you saw up there are the four solutions in terms of
 

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