complex number question help!!! (1 Viewer)

cormglakes · Mar 25, 2021

how do i do question ii:

and this question

ANS k = -4, 0

Drongoski · Mar 26, 2021

I'll just do part (i)

$\frac {1 + cos\theta+isin\theta}{1-cos\theta-isin\theta} = \frac {2cos^2\frac{\theta}{2}+2isin\frac{\theta}{2}cos\frac {\theta}{2}}{2sin^2\frac{\theta}{2}-2isin\frac{\theta}{2}cos\frac{\theta}{2}}= \frac {2cos\frac{\theta}{2}(cos\frac{\theta}{2}+isin\frac{\theta}{2})}{2sin\frac{\theta}{2}(sin\frac {\theta}{2} - icos \frac{\theta}{2})} \\ \\ = \frac{2cos\frac{\theta}{2}(i(sin\frac{\theta}{2} - icos\frac{\theta}{2}))}{2sin\frac{\theta}{2}(sin\frac{\theta}{2} - icos\frac{\theta}{2})}=icot\frac{\theta}{2}$

username_2 · Mar 26, 2021

cormglakes said:
how do i do question ii:
View attachment 30443
and this question
View attachment 30445

ANS k = 4, 0

CM_Tutor · Mar 27, 2021

NOTE: The results presented here contain a mistake corrected below and an approach predicated on the question not being flawed... see later posts for details.

username_2 said:
View attachment 30447

I got to the same point, but then ran into an issue.

$\begin{align*} \tan^8{\left(\frac{\theta}{2}\right)} &= -1 \\ &= \cos{\pi} + i\sin{\pi} \\ &= \cos{(2n\pi+\pi)} + i\sin{(2n\pi+\pi)} \qquad \text{for $n\in\mathbb{Z}$} \\ &= \cos{(2n+1)\pi} + i\sin{(2n+1)\pi} \\ \tan{\left(\frac{\theta}{2}\right)} &= \cos{\frac{(2n+1)\pi}{8}} + i\sin{\frac{(2n+1)\pi}{8}} \end{align*}$

Now, as $\theta$ must be real, the LHS must be real, and so we can conclude that:

$\begin{align*} \sin{\frac{(2n+1)\pi}{8}} &= 0 \\ \frac{(2n+1)\pi}{8} &= k\pi \qquad \text{for $k\in\mathbb{Z}$} \\ 2n+1 &= \frac{8}{\pi} \times k\pi \\ 2n &= 8k - 1 \end{align*}$

This equation has no solution as the LHS is even and the RHS is odd.

However, this reasoning requires $\tan{\left(\frac{\theta}{2}\right)}$ to be defined, which is untrue if $\theta = (2m+1)\pi \text{ for $m\in\mathbb{Z}$}$ .

In this case, $z= \cos\theta + i\sin\theta = \pm1$ .

Returning the original equation

$\left(\frac{z-1}{z+1}\right)^8 = -1$

it is clear that LHS is undefined for $z=-1$ and $z=1$ leads to LHS = 0.

Thus, there appear to be no solutions.

However, if start from the original equation:

$\begin{align*} \left(\frac{z-1}{z+1}\right)^8 &= -1 \\ (z-1)^8 &= -(z+1)^8 \\ (z-1)^8 + (z+1)^8 &= 0 \\ 2\left[\binom{8}{0} + \binom{8}{2}z^2 + \binom{8}{4}z^4 + \binom{8}{6}z^6 + \binom{8}{8}z^8\right] &= 0 \\ 1 + 28z^2 + 70z^4 + 28z^6 + z^8 &= 0 \\ \left(z^4 + \frac{1}{z^4}\right) + 28\left(z^2 + \frac{1}{z^2}\right) + 70 &= 0 \\ \left(z^2 + \frac{1}{z^2}\right)^2 - 2 + 28\left(z^2 + \frac{1}{z^2}\right) + 70 &= 0 \qquad \text{as $\left(z^2 + \frac{1}{z^2}\right)^2 = z^4 + \frac{1}{z^4} + 2$} \\ U^2 + 28U + 68 &= 0 \qquad \text{where $U = z^2 + \frac{1}{z^2}$} \\ U &= -14 \pm 8\sqrt{2} \\ z^2 + \frac{1}{z^2} &= 14 \pm 8\sqrt{2} \\ \left(z+\frac{1}{z}\right)^2 &= 16 \pm 8\sqrt{2} \end{align*}$

Let $4M_1 = 16 + 8\sqrt2 = 4\left(4+2\sqrt{2}\right) \implies M_1 = 4 + 2\sqrt{2} \text{ and } M_1 > 0$ and
let $4M_2 = 16 - 8\sqrt2 = 4\left(4-2\sqrt{2}\right) \implies M_2 = 4 - 2\sqrt{2} \text{ and } M_2 > 0$ .

We have now reduced the problem to two equations, $z + \frac{1}{z} = \pm 2\sqrt{M_1}$ and $z + \frac{1}{z} = \pm 2\sqrt{M_2}$

$\begin{align*} z+ \frac{1}{z} &= 2\sqrt{M} \\ z^2 + 1 &= 2z\sqrt{M} \\ z^2 - 2z\sqrt{M} + 1 &= 0 \\ z &= \frac{2\sqrt{M} \pm \sqrt{4M - 4(1)(1)}}{2} \\ &= \sqrt{M} \pm \sqrt{M - 1} \end{align*}$

This gives solutions $z = \sqrt{M_1} \pm \sqrt{M_1 - 1} = \sqrt{4 + 2\sqrt{2}} \pm \sqrt{3 + 2\sqrt{2}}$ and $z = \sqrt{M_2} \pm \sqrt{M_2 - 1} = \sqrt{4 - 2\sqrt{2}} \pm \sqrt{3 - 2\sqrt{2}}$

$\begin{align*} z+ \frac{1}{z} &= -2\sqrt{M} \\ z^2 + 1 &= -2z\sqrt{M} \\ z^2 + 2z\sqrt{M} + 1 &= 0 \\ z &= \frac{-2\sqrt{M} \pm \sqrt{4M - 4(1)(1)}}{2} \\ &= -\sqrt{M} \pm \sqrt{M - 1} \end{align*}$

This gives four more solutions $z = -\sqrt{M_1} \pm \sqrt{M_1 - 1} = -\sqrt{4 + 2\sqrt{2}} \pm \sqrt{3 + 2\sqrt{2}}$ and $z = -\sqrt{M_2} \pm \sqrt{M_2 - 1} = -\sqrt{4 - 2\sqrt{2}} \pm \sqrt{3 - 2\sqrt{2}}$

However, none of these solutions is valid as all are real and no real can have an eighth power that is negative.

Am I making some silly mistake, or is this really an elaborate "there are no solutions" problem... and where does the second method shown here break down, as it is giving 8 real answers but all are invalid, and I don't see the flaw in the method that produced them?

fan96 · Mar 27, 2021

CM_Tutor said:
Am I making some silly mistake, or is this really an elaborate "there are no solutions" problem... and where does the second method shown here break down, as it is giving 8 real answers but all are invalid, and I don't see the flaw in the method that produced them?

There appears to be a sign error here:

CM_Tutor said:
$\begin{align*} U &= -14 \pm 8\sqrt{2} \\ z^2 + \frac{1}{z^2} &= 14 \pm 8\sqrt{2} \end{align*}$

(The first attempt at a solution fails because it assumes all solutions to part ii) lie on the unit circle.)

fan96 · Mar 27, 2021

One possible solution:

Define

$f(z) = \frac{z-1}{z+1},$

$g(z) = \frac{1+z}{1-z}.$

Note that these functions are inverses when $z \ne -1, 1$ .

Let $w_n = \exp(i \cdot ((2n+1)\pi)/8)$ be the solutions to $w^8 = -1$ .

Now if

$\left(\frac{z-1}{z+1} \right)^8 = f(z)^8 = -1,$

then we know that any solution $z$ satisfies $f(z) = w_n$ for some $n$ .

Apply $g$ to both sides, so we get $z = g(w_n).$

Because we have an explicit formula for $g$ we're now done, but we can use part i) to get a nicer form for the solutions.

From part i), we know that $g(z) = i \cot(\theta/2)$ for any $z = \exp(i\theta)$ on the unit circle.
(Geometrically, this means that $g$ sends the unit circle to the imaginary axis. In other words, all our solutions will be purely imaginary.)

From here, we just need to evaluate $g(w_n) = i \cot((2n+1)\pi)/16).$

The co-tangent function has periodicity $\pi$ , so take $n = 0 ... 7$ to obtain the full set of solutions.

Pedantic note: We show here that every solution $z$ corresponds to a $g(w_n)$ for some $n$ .
Technically, we should also explain why every $g(w_n)$ is a valid solution $z$ .
This is because there are exactly eight possible values each for $z$ and $w_n$ .
None of these values are $1$ or $-1$ , so the function $f$ is a one-to-one mapping between these sets.

---

Fun fact: $f$ and $g$ are examples of Möbius transformations - transformations of the type

$z \mapsto \frac{az+b}{cz+d}, \quad a,b,c,d \in \mathbb R.$

Such transformations in the complex plane send a line or circle to another line or circle.

CM_Tutor · Mar 28, 2021

fan96 said:
There appears to be a sign error here:

(The first attempt at a solution fails because it assumes all solutions to part ii) lie on the unit circle.)

Thanks for pointing out the sign error, I will come back and post the corrected solution... we now have all the cases with $z^2+z^{-2}<0$ leading to only non-real complex solutions, as was expected.

CM_Tutor · Mar 28, 2021

Returning the original equation and correcting my earlier mistake, to find the actual 8 solutions of the equation:

$\begin{align*} \left(\frac{z-1}{z+1}\right)^8 &= -1 \\ (z-1)^8 &= -(z+1)^8 \\ (z-1)^8 + (z+1)^8 &= 0 \\ 2\left[\binom{8}{0} + \binom{8}{2}z^2 + \binom{8}{4}z^4 + \binom{8}{6}z^6 + \binom{8}{8}z^8\right] &= 0 \\ 1 + 28z^2 + 70z^4 + 28z^6 + z^8 &= 0 \\ \left(z^4 + \frac{1}{z^4}\right) + 28\left(z^2 + \frac{1}{z^2}\right) + 70 &= 0 \\ \left(z^2 + \frac{1}{z^2}\right)^2 - 2 + 28\left(z^2 + \frac{1}{z^2}\right) + 70 &= 0 \qquad \text{as $\left(z^2 + \frac{1}{z^2}\right)^2 = z^4 + \frac{1}{z^4} + 2$} \\ U^2 + 28U + 68 &= 0 \qquad \text{where $U = z^2 + \frac{1}{z^2}$} \\ U &= -14 \pm 8\sqrt{2} \\ z^2 + \frac{1}{z^2} &= -14 \pm 8\sqrt{2} \\ \left(z+\frac{1}{z}\right)^2 &= -12 \pm 8\sqrt{2} \end{align*}$

Let $4M_1 = -12 + 8\sqrt2 = 4\left(-3+2\sqrt{2}\right) \implies M_1 = 2\sqrt{2} - 3 \text{ and } M_1 < 0$ and
let $4M_2 = -12 - 8\sqrt2 = 4\left(-3-2\sqrt{2}\right) \implies M_2 = -3 - 2\sqrt{2} \text{ and } M_2 < 0$ .

With both $M_1$ and $M_2$ being negative, we can be certain that $z + z^{-1}$ are non-real.

Further, we can rewrite $M_1$ as:

$-3 + 2\sqrt{2} = -\left(1 - 2\sqrt{2} + 2\right) = -\left[1^2 - 2(1)\left(\sqrt{2}\right) + \left(\sqrt{2}\right)^2\right] = i^2\left(1-\sqrt{2}\right)^2$

and rewrite $M_2$ as:

$-3 - 2\sqrt{2} = -\left(1 + 2\sqrt{2} + 2\right) = -\left[1^2 + 2(1)\left(\sqrt{2}\right) + \left(\sqrt{2}\right)^2\right] = i^2\left(1+\sqrt{2}\right)^2$

We have now reduced the problem to two equations, $z + \frac{1}{z} = \pm 2\sqrt{M_1} = \pm 2i\left(1-\sqrt{2}\right)$ and $z + \frac{1}{z} = \pm 2\sqrt{M_2} = \pm 2i\left(1+\sqrt{2}\right)$

$\begin{align*} \text{For any $k \in \mathbb{R}$} \qquad z+ \frac{1}{z} &= 2ik \\ z^2 + 1 &= 2ikz \\ z^2 - 2ikz + 1 &= 0 \\ z &= \frac{2ik \pm \sqrt{4i^2k^2 - 4(1)(1)}}{2} \\ &= \frac{2ik \pm \sqrt{4i^2k^2 + 4i^2}}{2} \\ &= \frac{2ik \pm 2i\sqrt{k^2 + 1}}{2} \\ &= i\left(k \pm \sqrt{k^2 + 1}\right) \end{align*}$

Our eight solutions thus occur when

$\begin{align*} k &= 1 - \sqrt{2} \text{, in which case } k^2 = 3 - 2\sqrt{2} \implies k^2 + 1 = 4 - 2\sqrt{2} \\ k &= \sqrt{2} - 1 \text{, in which case } k^2 = 3 - 2\sqrt{2} \implies k^2 + 1 = 4 - 2\sqrt{2} \\ k &= 1 + \sqrt{2} \text{, in which case } k^2 = 3 + 2\sqrt{2} \implies k^2 + 1 = 4 + 2\sqrt{2} \\ k &= -1 - \sqrt{2} \text{, in which case } k^2 = 3 + 2\sqrt{2} \implies k^2 + 1 = 4 + 2\sqrt{2} \end{align*}$

and so are:

$z = i\left(1 - \sqrt{2} \pm \sqrt{4 - 2\sqrt{2}}\right)$

$z = i\left(\sqrt{2} - 1 \pm \sqrt{4 - 2\sqrt{2}}\right)$

$z = i\left(1 + \sqrt{2} \pm \sqrt{4 + 2\sqrt{2}}\right)$

$z = i\left(-1 - \sqrt{2} \pm \sqrt{4 + 2\sqrt{2}}\right)$

CM_Tutor · Mar 28, 2021

cormglakes said:
how do i do question ii:
View attachment 30443

This first question is flawed. Putting $z=\cos\theta + i\sin\theta$ does yield the result $\frac{z-1}{z+1}=i\cot{\frac{\theta}{2}}$ . Applying it to solve the second equation, however, is only valid if the second part of the equation has $z$ restricted to the unit circle and thus having a modulus of 1. Looking at the eight solutions that I have found algebraically, none of them has $|z|=1$ and so none can be found using the result in part (i).

So, there is no "hence" solution possible here and this is a textbook mistake. Where does this come from, @cormglakes?

CM_Tutor · Mar 28, 2021

Adapting from @fan96 's approach, we can demonstrate that my 8 solutions correspond to solutions in the form of $i\cot\theta$ and so can be used to prove results like:

$\cot{\frac{\pi}{16}} = 1+\sqrt{2}+\sqrt{4+2\sqrt{2}}$

fan96 · Mar 28, 2021

CM_Tutor said:
This first question is flawed. Putting $z=\cos\theta + i\sin\theta$ does yield the result $\frac{z-1}{z+1}=i\cot{\frac{\theta}{2}}$ . Applying it to solve the second equation, however, is only valid if the second part of the equation has $z$ restricted to the unit circle and thus having a modulus of 1. Looking at the eight solutions that I have found algebraically, none of them has $|z|=1$ and so none can be found using the result in part (i).

So, there is no "hence" solution possible here and this is a textbook mistake. Where does this come from, @cormglakes?

In my solution, the $z$ for which the substitution $\cos \theta + i \sin \theta$ occurs are the eighth roots of $-1$ , not the actual solutions of ii). This is a valid substitution, just not the typical one that would be expected here.

CM_Tutor · Mar 28, 2021

fan96 said:
In my solution, the $z$ for which the substitution $\cos \theta + i \sin \theta$ occurs are the eighth roots of $-1$ , not the actual solutions of ii). This is a valid substitution, just not the typical one that would be expected here.

You are correct, of course, and your method is valid as those all satisfy $|z|=1$ .

I was explaining why the approach implied by the question that leads to

$\tan^8{\left(\frac{\theta}{2}\right)}=-1$

is an invalid approach because it supposes that the roots of

$\left(\frac{z-1}{z+1}\right)^8 = -1$

have the property that $|z|=1$ and they don't.

CM_Tutor · Mar 29, 2021

cormglakes said:
View attachment 30445

ANS k = 4, 0

If $z=w$ is a root of $z^3 + iz^2 + ikz + 2i = 0$ where $k\in\mathbb{R}$ then $w^3 + iw^2 + ikw + 2i = 0$ .

Since $(1-i)w \in \mathbb{R}$ , let $m$ be the real number equal to $(1-i)w$ , and so

$\begin{align*} (1-i)w &= m \\ w &= \frac{m}{1-i} \times \frac{1+i}{1+i} \\ &= \frac{m(1+i)}{1^2 - i^2} \\ &= \frac{m}{2}(1 + i) \end{align*}$

Substituting this value for $w$ into the above equation yields:

$\begin{align*} \left[\frac{m}{2}(1 + i)\right]^3 + i\left[\frac{m}{2}(1 + i)\right]^2 + ik\left[\frac{m}{2}(1 + i)\right] + 2i &= 0 \\ \frac{m^3}{8}(1 + 3i + 3i^2 + i^3) + \frac{im^2}{4}(1 + 2i + i^2) + \frac{ikm}{2}(1 + i) + 2i &= 0 \\ \frac{m^3}{2}(-2 + 2i) + im^2(2i) + 2ikm(1 + i) + 8i &= 0 \\ -m^3 + im^3 + 2i^2m^2 + 2ikm + 2i^2km + 8i &= 0 \\ -m^3 - 2m^2 - 2km + i(m^3 + 2km + 8) &= 0 \end{align*}$

Equating the real and imaginary parts gives two equations:

$-m^3 - 2m^2 - 2km = 0 \qquad \text{. . . . . . . . .(1)}$

and

$m^3 + 2km + 8 = 0 \qquad \text{. . . . . . . . .(2)}$

From (1), we see that $m^3 + 2km = -2m^2$ which can be put into (2) to yield $-2m^2 + 8 = 0 \implies m^2 = 4 \implies m = \pm2$

Making $k$ the subject of (2) yields $k = \frac{-(m^3 + 8)}{2m}$ and thus:

$\text{If $m = +2$:} \quad k = \frac{-(m^3 + 8)}{2m} = \frac{-(2^3 + 8)}{2 \times 2} = \frac{-16}{4} = -4$

$\text{If $m = -2$:} \quad k = \frac{-(m^3 + 8)}{2m} = \frac{-[(-2)^3 + 8]}{2 \times -2} = \frac{0}{-4} = 0$

So, there are two solutions, $k = -4$ and $k = 0$ .

@cormglakes, can you check if your $k = 4$ answer is a typo?

cormglakes · Apr 10, 2021

CM_Tutor said:
If $z=w$ is a root of $z^3 + iz^2 + ikz + 2i = 0$ where $k\in\mathbb{R}$ then $w^3 + iw^2 + ikw + 2i = 0$ .

Since $(1-i)w \in \mathbb{R}$ , let $m$ be the real number equal to $(1-i)w$ , and so

$\begin{align*} (1-i)w &= m \\ w &= \frac{m}{1-i} \times \frac{1+i}{1+i} \\ &= \frac{m(1+i)}{1^2 - i^2} \\ &= \frac{m}{2}(1 + i) \end{align*}$

Substituting this value for $w$ into the above equation yields:

$\begin{align*} \left[\frac{m}{2}(1 + i)\right]^3 + i\left[\frac{m}{2}(1 + i)\right]^2 + ik\left[\frac{m}{2}(1 + i)\right] + 2i &= 0 \\ \frac{m^3}{8}(1 + 3i + 3i^2 + i^3) + \frac{im^2}{4}(1 + 2i + i^2) + \frac{ikm}{2}(1 + i) + 2i &= 0 \\ \frac{m^3}{2}(-2 + 2i) + im^2(2i) + 2ikm(1 + i) + 8i &= 0 \\ -m^3 + im^3 + 2i^2m^2 + 2ikm + 2i^2km + 8i &= 0 \\ -m^3 - 2m^2 - 2km + i(m^3 + 2km + 8) &= 0 \end{align*}$

Equating the real and imaginary parts gives two equations:

$-m^3 - 2m^2 - 2km = 0 \qquad \text{. . . . . . . . .(1)}$

and

$m^3 + 2km + 8 = 0 \qquad \text{. . . . . . . . .(2)}$

From (1), we see that $m^3 + 2km = -2m^2$ which can be put into (2) to yield $-2m^2 + 8 = 0 \implies m^2 = 4 \implies m = \pm2$

Making $k$ the subject of (2) yields $k = \frac{-(m^3 + 8)}{2m}$ and thus:

$\text{If $m = +2$:} \quad k = \frac{-(m^3 + 8)}{2m} = \frac{-(2^3 + 8)}{2 \times 2} = \frac{-16}{4} = -4$

$\text{If $m = -2$:} \quad k = \frac{-(m^3 + 8)}{2m} = \frac{-[(-2)^3 + 8]}{2 \times -2} = \frac{0}{-4} = 0$

So, there are two solutions, $k = -4$ and $k = 0$ .

@cormglakes, can you check if your $k = 4$ answer is a typo?

sorry yea the answer is -4,0

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