Help with These 5 complex number questions (1 Viewer)

MAK03 · Jan 3, 2022

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

jimmysmith560 · Jan 3, 2022

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

The working to the above question is as follows:

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!

5uckerberg · Jan 3, 2022

With the fourth image, the question you gonna ask yourself is how can we get from $\tan{\alpha}$ to $\tan{4\alpha}$ . 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
$\tan(\alpha+\beta)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}$ . 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

$\tan(\alpha+\alpha)=\frac{\tan{\alpha}+\tan{\alpha}}{1-\tan{\alpha}\tan{\alpha}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4}$

Note one can say that $\tan{2\alpha}=\frac{2\tan{\alpha}}{1-\tan^{2}{\alpha}}$

Next step

$\tan(4\alpha)=\frac{\tan{2\alpha}+\tan{2\alpha}}{1-\tan{2\alpha}\tan{2\alpha}}=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{24}{7}$

There you will obtain $\alpha=tan^{-1}\frac{24}{7}$

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 $\theta$ on its own means a restriction of $0 < \alpha < \frac{\pi}{2}$ . But remember what we did in part a we went into the realm of $4\alpha$ so therefore our restriction instead should be

$0 < \alpha < 2\pi$

Remember what you have here is to the fourth power which gives the length of 25 so thus your r should be $\sqrt{5}$ . The angles is a whole different story. What we have to do is find $\tan^{-1}{\frac{24}{7}}$ and for 4 cycles okay so we have to add $2\pi$ on that so therefore what we will obtain is
$\sqrt{5}(\cos{\frac{\tan^{-1}{\frac{24}{7}}}{4}+isin{\frac{\tan^{-1}{\frac{24}{7}}}{4})[TEX] and [TEX]\sqrt{5}(\cos{\frac{\tan^{-1}{\frac{24}{7}}+2\pi}{4}+isin{\frac{\tan^{-1}{\frac{24}{7}}+2\pi}{4})$
and
$\sqrt{5}(\cos{\frac{\tan^{-1}{\frac{24}{7}}+4\pi}{4}+isin{\frac{\tan^{-1}{\frac{24}{7}}+4\pi}{4})$
and
$\sqrt{5}(\cos{\frac{\tan^{-1}{\frac{24}{7}}+6\pi}{4}+isin{\frac{\tan^{-1}{\frac{24}{7}}+6\pi}{4})$

As specified the values of $\theta$ are found here
$\frac{\tan^{-1}{\frac{24}{7}}}{4}$
$\frac{\tan^{-1}{\frac{24}{7}}+2\pi}{4}$
$\frac{\tan^{-1}{\frac{24}{7}}+4\pi}{4}$
$\frac{\tan^{-1}{\frac{24}{7}}+6\pi}{4}$

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]

5uckerberg · Jan 3, 2022

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.

Lith_30 · Jan 3, 2022

For image 1:

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

part ii)
$\\\alpha+\gamma=1+\beta\\\alpha^2+2\alpha\gamma+\gamma^2=(1+\beta)^2\\\alpha^2-2\alpha\gamma+\gamma^2=(1+\beta)^2-4\left(\frac{1}{2}(1+\beta^2)\right)\ \text{from the identity derived from the product of roots}\\\\(\alpha-\gamma)^2=1+2\beta+\beta^2-2-2\beta^2\\(\alpha-\gamma)^2=-(\beta^2-2\beta+1)\\(\alpha-\gamma)^2=-(1-\beta)^2\\\alpha-\gamma=\pm{i}(1-\beta)\ \text{as required}$

part iii)
Not too sure about this, but by using the identity $\alpha-\gamma=\pm{i}(1-\beta)$ 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 $\beta=1+i$ .

we can sub beta into $\alpha-\gamma=\pm{i}(1-\beta)$
$\\\alpha-\gamma=\pm{i}(1-(1+i))\\\alpha-\gamma=\pm{1}$
we can then add this expression onto $\alpha+\gamma=1+\beta$
$\\2\alpha=1+(1+i)\pm{1}\\\alpha=\frac{1}{2}(2\pm{1}+i)$

Therefore the other two roots of the equation are $\frac{1}{2}+\frac{1}{2}i$ and $\frac{3}{2}+\frac{1}{2}i$

5uckerberg · Jan 3, 2022

The first question in part i is effectively asking whether $\beta$ and 1 are the roots of the equation. FIrst and foremost sub in $x=1, x=\beta$ . 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 $\frac{-b}{a}$ and product of roots for a polynomial of degree 4 is $\frac{e}{a}$ for any given polynomial of the form $ax^{4}+bx^{3}+cx^{2}+dx+e$ .

As noted the sum of the roots gives $\frac{-2(1+\beta)}{1}$ which if noted clearly b is the value $2(1+\beta)$
and using what we know $b=1+\beta+\alpha+\gamma$ , thus $\alpha+\gamma=1+\beta$ .

The product of 4 roots gives us $\frac{1}{2}\beta(1+\beta)^{2}$ . We know the product of $1 \times \beta = \beta$ so therefore the product should give us $1 \times \beta \times \alpha \times \gamma = \frac{1}{2}\beta(1+\beta)^{2}$ . Therefore, $\alpha\gamma = \frac{1}{2}(1+\beta^{2})$ .

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
$(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta$ . Thus, for this question this idea becomes $(\alpha-\gamma)^{2}=(\alpha+\gamma)^{2}-4\alpha\gamma$ .

There, $(\alpha-\gamma)^{2}=(1+\beta)^{2}-2(1+\beta^{2})=-\beta^{2}+2\beta-1=-(1-\beta)^{2}$ .

My next question to you is how do you find $(\alpha-\gamma)^{2} = -(1-\beta)^{2}$ .
Or should I say how do you find the square root of a negative number

MAK03 · Jan 3, 2022

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

5uckerberg · Jan 4, 2022

For question 2 we are given $z=1+\cos{\theta}+i\sin{\theta}$

The modulus requires us to find $\sqrt{(1+\cos{\theta})^{2}+\sin^{2}\theta} = \sqrt{2+2\cos{\theta}}$ . Now notice that you need to use the double angle formula to notice that $\cos{\frac{\theta}{2}}=\sqrt{\frac{\cos{\theta}+1}{2}}$ . Multiply by 2 and you will have $2\cos{\frac{\theta}{2}}=2\sqrt{\frac{\cos{\theta}+1}{2}}$ . Now 2 is just the square root of 4 right, so therefor multiply the RHS your modulus is just $2\cos{\frac{\theta}{2}}$ .

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 $modulus(\cos{arg}+i\sin{\arg})$ . Remember we have
$2\cos{\frac{\theta}{2}}$ as our modulus.

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

See here $\sin{2\theta}=2\sin{\theta}\cos{\theta}$
Well in that case you can clearly see that
$\sin{\theta}=2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}$ .
Using the double angle formula
$\cos{2\theta}=2\cos^{2}\theta-1$
then it is obvious that
$\cos{\theta}=2\cos^{2}\frac{\theta}{2}-1$
and $\cos{\theta}+1=2\cos^{2}\frac{\theta}{2}$
now if we are finding the mod arg form then it becomes
$2\cos{\frac{\theta}{2}}(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}})$
Thus your final answer will be that the argument is $\frac{\theta}{2}$

This part simply asks you to do this convert $(1+\cos{\theta}+i\sin{\theta})^{n}$ 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
$(1+\cos{\theta})^{4}=1+4\cos{\theta}+6\cos^{2}{\theta}+4\cos^{3}{\theta}+\cos^{4}{\theta}$ . As noticed for the modulus argument form we will also take the real part of the complex number giving us $\left(2\cos{\frac{\theta}{2}}\right)^{4}$ . Now as noted the modulus part when you raise to the power of 4 it gives us $16\cos^{4}{\frac{\theta}{2}}$ and for the argument part we use De Moivre's Theorem which will give us $\cos\left({\frac{\theta\times{4}}{2}}\right) = \cos{2\theta}$ . Once you have that you can clearly see that from the question $(1+\cos{\theta})^{4} = 16\cos^{4}{\frac{\theta}{2}} \cos{2\theta}$
giving us $1+4\cos{\theta}+6\cos^{2}{\theta}+4\cos^{3}{\theta}+\cos^{4}{\theta}=16\cos^{4}{\frac{\theta}{2}} \cos{2\theta}$ .

5uckerberg · Jan 4, 2022

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 · Jan 12, 2022

5uckerberg said:
For question 2 we are given $z=1+\cos{\theta}+i\sin{\theta}$

The modulus requires us to find $\sqrt{(1+\cos{\theta})^{2}+\sin^{2}\theta} = \sqrt{2+2\cos{\theta}}$ . Now notice that you need to use the double angle formula to notice that $\cos{\frac{\theta}{2}}=\sqrt{\frac{\cos{\theta}+1}{2}}$ . Multiply by 2 and you will have $2\cos{\frac{\theta}{2}}=2\sqrt{\frac{\cos{\theta}+1}{2}}$ . Now 2 is just the square root of 4 right, so therefor multiply the RHS your modulus is just $2\cos{\frac{\theta}{2}}$ .

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 $modulus(\cos{arg}+i\sin{\arg})$ . Remember we have
$2\cos{\frac{\theta}{2}}$ as our modulus.

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

See here $\sin{2\theta}=2\sin{\theta}\cos{\theta}$
Well in that case you can clearly see that
$\sin{\theta}=2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}$ .
Using the double angle formula
$\cos{2\theta}=2\cos^{2}\theta-1$
then it is obvious that
$\cos{\theta}=2\cos^{2}\frac{\theta}{2}-1$
and $\cos{\theta}+1=2\cos^{2}\frac{\theta}{2}$
now if we are finding the mod arg form then it becomes
$2\cos{\frac{\theta}{2}}(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}})$
Thus your final answer will be that the argument is $\frac{\theta}{2}$

This part simply asks you to do this convert $(1+\cos{\theta}+i\sin{\theta})^{n}$ 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
$(1+\cos{\theta})^{4}=1+4\cos{\theta}+6\cos^{2}{\theta}+4\cos^{3}{\theta}+\cos^{4}{\theta}$ . As noticed for the modulus argument form we will also take the real part of the complex number giving us $\left(2\cos{\frac{\theta}{2}}\right)^{4}$ . Now as noted the modulus part when you raise to the power of 4 it gives us $16\cos^{4}{\frac{\theta}{2}}$ and for the argument part we use De Moivre's Theorem which will give us $\cos\left({\frac{\theta\times{4}}{2}}\right) = \cos{2\theta}$ . Once you have that you can clearly see that from the question $(1+\cos{\theta})^{4} = 16\cos^{4}{\frac{\theta}{2}} \cos{2\theta}$
giving us $1+4\cos{\theta}+6\cos^{2}{\theta}+4\cos^{3}{\theta}+\cos^{4}{\theta}=16\cos^{4}{\frac{\theta}{2}} \cos{2\theta}$ .

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

5uckerberg · Jan 12, 2022

MAK03 said:
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
$\left(1+\cos{\theta}+i\sin{\theta}\right)^{4}=\left(1+\cos{\theta}\right)^{4}-6\sin^{2}{\theta}\left(1+\cos{\theta}\right)^{2}+\sin^{4}{\theta}$ .

Simplifying we will have
$\left(1+\cos{\theta}\right)^{4}-6\left(1+\cos{\theta}\right)^{2}+6\cos^{2}{\theta}\left(1+\cos{\theta}\right)^{2}+\left(1-\cos^{2}{\theta}\right)^{2}$

Expanding will obtain
$1+4\cos{\theta}+6\cos^{2}{\theta}+4\cos^{3}{\theta}+\cos^{4}{\theta}-6\left(1+2\cos{\theta}+\cos^{2}{\theta}\right)+6\cos^{2}{\theta}\left(1+2\cos{\theta}+\cos^{2}{\theta}\right)+1-2\cos^{2}{\theta}+\cos^{4}{\theta}$

Collecting all the terms we are then left with
$8\cos^{4}{\theta}+16\cos^{3}{\theta}+4\cos^{2}{\theta}-8\cos{\theta}-4$

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 $z+z^{-1}=2\cos{\theta}$ we can now play with this problem

$8\cos^{4}{\theta}+16\cos^{3}{\theta}+4\cos^{2}{\theta}-8\cos{\theta}-4$ is simply
$\left(\frac{(z+z^{-1})^{4}}{2}\right)+2\left(z+z^{-1}\right)^{3}+\left(z+z^{-1}\right)^{2}-4\left(z+z^{-1}\right)-4$

Expanding that we will have

$\left(\frac{z^{4}+4z^{2}+6+4z^{-2}+z^{-4}}{2}\right)+2\left(z^{3}+3z+3z^{-1}+z^{-3}\right)+z^{2}+2+z^{-2}-8\cos{\theta}-4$

Part 2 will come

5uckerberg · Jan 12, 2022

Which simplified itself will be
$\left(\cos{4\theta}+4\cos{2\theta}+3\right)+4\cos{3\theta}+12\cos{\theta}+2\cos{2\theta}+2-8\cos{\theta}-4$

Simplifying further gives us
$\cos{4\theta}+4\cos{3\theta}+6\cos{2\theta}+4\cos{\theta}+1$

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 $\left(\frac{z-z^{-1}}{i}\right)=2\sin{\theta}$

5uckerberg · Jan 13, 2022

What we should have is $4i\left(1+\cos{\theta}\right)^{3}\sin{\theta}-4i\left(1+\cos{\theta}\right)\sin^{3}{\theta}$ from the binomial expansion

Expanding it becomes

$4i\sin{\theta}\left(1+3\cos{\theta}+3\cos^{2}{\theta}+\cos^{3}{\theta}\right)-4i\sin^{3}{\theta}-4i\sin^{3}{\theta}\cos{\theta}$

$4i\sin{\theta}\cos^{3}{\theta}-4i\sin^{3}{\theta}\cos{\theta}-4i\sin^{3}{\theta}+12i\sin{\theta}\cos^{2}{\theta}+12i\sin{\theta}\cos{\theta}+4i\sin{\theta}$

Notice when I said $\left(z-z^{-1}\right)=2i\sin{\theta}$

It will become $\left(z-z^{-1}\right)2\cos^{3}{\theta}-\frac{\left(z-z^{-1}\right)^{3}}{2}\cos{\theta}-\frac{\left(z-z^{-1}\right)^{3}}{2}+\left(z-z^{-1}\right)6\cos^{2}{\theta}+\left(z-z^{-1}\right)6\cos{\theta}+2\left(z-z^{-1}\right)$

Now do you recall this $z+z^{-1}=2\cos{\theta}$

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

$\left(z-z^{-1}\right)\frac{\left(z+z^{-1}\right)^{3}}{4}-\frac{\left(z-z^{-1}\right)^{3}}{2}\frac{\left(z+z^{-1}\right)}{2}-\frac{\left(z-z^{-1}\right)^{3}}{2}+\left(z-z^{-1}\right)\frac{3\left(z+z^{-1}\right)^{2}}{2}+3\left(z-z^{-1}\right)\left(z+z^{-1}\right)+2\left(z-z^{-1}\right)$

Simplifying this will give us

$\frac{\left(z-z^{-1}\right)\left(z+z^{-1}\right)^{3}}{4}-\frac{\left(z-z^{-1}\right)^{3}\left(z+z^{-1}\right)}{4}-\frac{\left(z-z^{-1}\right)^{3}}{2}+\frac{3\left(z-z^{-1}\right)\left(z+z^{-1}\right)^{2}}{2}+3\left(z-z^{-1}\right)\left(z+z^{-1}\right)+2\left(z-z^{-1}\right)$

5uckerberg · Jan 13, 2022

Which can be factorised as

$\frac{\left(z-z^{-1}\right)\left(z+z^{-1}\right)}{4}=\frac{z^{2}-z^{-2}}{4}$
$\frac{z^{2}-z^{-2}}{4}\left(\left(z+z^{-1}\right)^{2}-\left(z-z^{-1}\right)^{2}\right)-\frac{\left(z^{3}-3z+3z^{-1}-z^{-3}\right)}{2}+\frac{3\left(z-z^{-1}\right)\left(z^{2}+2+z^{-2}\right)}{2}+3\left(z^{2}-z^{-2}\right)+2\left(z-z^{-1}\right)$

This is just $\frac{\left(z^{2}-z^{-2}\right)\left(z+z^{-1}\right)}{2}-\frac{\left(z^{3}-3z+3z^{-1}-z^{-3}\right)}{2}+\frac{3\left(z^{3}-z^{-3}+z-z^{-1}\right)}{2}+3\left(z^{2}-z^{-2}\right)+2\left(z-z^{-1}\right)$

Further simplification will give us

$2\left(z^{3}-z^{-3}+z-z^{-1}\right)-\frac{\left(z^{3}-3z+3z^{-1}-z^{-3}\right)}{2}+3\left(z^{2}-z^{-2}\right)+2\left(z-z^{-1}\right)$

Later on, it becomes $\frac{3}{2}\left(z^{3}-z^{-3}\right)+3\left(z^{2}-z^{-2}\right)+\frac{11}{2}\left(z-z^{-1}\right)$

MAK03 · Jan 17, 2022

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

MAK03 · Jan 19, 2022

5uckerberg said:
With the fourth image, the question you gonna ask yourself is how can we get from $\tan{\alpha}$ to $\tan{4\alpha}$ . 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
$\tan(\alpha+\beta)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}$ . 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

$\tan(\alpha+\alpha)=\frac{\tan{\alpha}+\tan{\alpha}}{1-\tan{\alpha}\tan{\alpha}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4}$

Note one can say that $\tan{2\alpha}=\frac{2\tan{\alpha}}{1-\tan^{2}{\alpha}}$

Next step

$\tan(4\alpha)=\frac{\tan{2\alpha}+\tan{2\alpha}}{1-\tan{2\alpha}\tan{2\alpha}}=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{24}{7}$

There you will obtain $\alpha=tan^{-1}\frac{24}{7}$

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 $\theta$ on its own means a restriction of $0 < \alpha < \frac{\pi}{2}$ . But remember what we did in part a we went into the realm of $4\alpha$ so therefore our restriction instead should be

$0 < \alpha < 2\pi$

Remember what you have here is to the fourth power which gives the length of 25 so thus your r should be $\sqrt{5}$ . The angles is a whole different story. What we have to do is find $\tan^{-1}{\frac{24}{7}}$ and for 4 cycles okay so we have to add $2\pi$ on that so therefore what we will obtain is
$\sqrt{5}(\cos{\frac{\tan^{-1}{\frac{24}{7}}}{4}+isin{\frac{\tan^{-1}{\frac{24}{7}}}{4})[TEX] and [TEX]\sqrt{5}(\cos{\frac{\tan^{-1}{\frac{24}{7}}+2\pi}{4}+isin{\frac{\tan^{-1}{\frac{24}{7}}+2\pi}{4})$
and
$\sqrt{5}(\cos{\frac{\tan^{-1}{\frac{24}{7}}+4\pi}{4}+isin{\frac{\tan^{-1}{\frac{24}{7}}+4\pi}{4})$
and
$\sqrt{5}(\cos{\frac{\tan^{-1}{\frac{24}{7}}+6\pi}{4}+isin{\frac{\tan^{-1}{\frac{24}{7}}+6\pi}{4})$

As specified the values of $\theta$ are found here
$\frac{\tan^{-1}{\frac{24}{7}}}{4}$
$\frac{\tan^{-1}{\frac{24}{7}}+2\pi}{4}$
$\frac{\tan^{-1}{\frac{24}{7}}+4\pi}{4}$
$\frac{\tan^{-1}{\frac{24}{7}}+6\pi}{4}$

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 · Jan 20, 2022

MAK03 said:
Hey again, could you please re-explain part b and c

So for part b start with the fact that the length of $7+24i$ is 25. and then the r which is length is the fourth root of 25 equalling $\sqrt{5}$ . The argument is very simple as well. First we see that for $\cos{\theta}+i\sin{\theta}$ is $\cos\left({\tan^{-1}\frac{24}{7}+n\pi}\right)+i\sin\left({\tan^{-1}\frac{24}{7}+n\pi}\right)$ where $n=0, 1, -1, -2$ so therefore, your four values are $\sqrt{5}\left(\cos\left({4\alpha+\pi}\right)+i\sin\left({4\alpha+\pi}\right)\right), \sqrt{5}\left(\cos\left({4\alpha}\right)+i\sin\left({4\alpha}\right)\right), \sqrt{5}\left(\cos\left({4\alpha-\pi}\right)+i\sin\left({4\alpha-\pi}\right)\right), \sqrt{5}\left(\cos\left({4\alpha-2\pi}\right)+i\sin\left({4\alpha-2\pi}\right)\right)$ the last statement is simply just $\sqrt{5}\left(\cos\left({4\alpha}\right)+i\sin\left({4\alpha}\right)\right)$ because by taking away $2\pi$ you are back to where you started.

Why is this working so well, remember from the question it specified that "Given that $0 < \alpha < \frac{\pi}{2} ...$ and that one has to express 7+24i in the form $r\left(\cos{\theta}+i\sin{\theta}\right)$ then $\theta$ has to be $\tan^{-1}\frac{24}{7}$ which is simply $4\alpha$ as such $\theta=4\alpha$ and there the boundaries suddenly became $0 < \theta < 2\pi$ so therefore, what you saw up there are the four solutions in terms of $\alpha$

5uckerberg · Jan 20, 2022

For part c the answer that comes to me is just $\frac{3}{\sqrt{2}}+\frac{i}{\sqrt{2}}, -\left(\frac{3}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right), \frac{1}{\sqrt{2}}+\frac{3i}{\sqrt{2}}, -\left(\frac{1}{\sqrt{2}}+\frac{3i}{\sqrt{2}}\right)$

Your four roots in the form a+ib @MAK03

Help with These 5 complex number questions (1 Viewer)

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