0
$\begingroup$

Question: If $$z = \left(\frac{\sqrt3}{2} + \frac{i}{2}\right)^{107} + \left(\frac{\sqrt3}{2} - \frac{i}{2}\right)^{107} $$ Show that Im(z) = 0

I have no idea how to even start the question. Please point me in the right direction.

$\endgroup$
  • 1
    $\begingroup$ Have you tried writing the question in exponential form? $\endgroup$ – Peter Woolfitt Jul 27 '14 at 5:51
  • $\begingroup$ @PeterWoolfitt We haven't learnt the exponential form yet. $\endgroup$ – Gummy bears Jul 27 '14 at 6:08
  • $\begingroup$ @Gummybears Have you seen binomial expansion? I added a hint with more detail. $\endgroup$ – Peter Woolfitt Jul 27 '14 at 6:35
3
$\begingroup$

Hint: second term is the complex conjugate of the first..

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes I realized that. But don't know how to proceed next :( $\endgroup$ – Gummy bears Jul 27 '14 at 6:04
  • $\begingroup$ Use the fact that $\bar{z}^\alpha = \overline{z^\alpha}$. Then what can you say about $w + \bar{w}$? $\endgroup$ – LinAlgMan Jul 27 '14 at 6:36
  • $\begingroup$ That they cancel out the imaginary part. But what can I do about the power? @LinAlgMan $\endgroup$ – Gummy bears Jul 27 '14 at 7:03
  • $\begingroup$ See the begining of my comment. $\endgroup$ – LinAlgMan Jul 27 '14 at 7:31
2
$\begingroup$

\begin{align} z &=\left(\frac{\sqrt3}{2} + \frac{i}{2}\right)^{107} + \left(\frac{\sqrt3}{2} - \frac{i}{2}\right)^{107}\\ &=2\cdot\frac{e^{107i\pi/6}+e^{-107i\pi/6}}{2}\\ &=2\cos\left(\frac{107\pi}{6}\right)\\ &=\sqrt 3 \end{align} Thus $\Im (z)=0$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Great! The Euler form. Why didn't I thought of this? $\endgroup$ – Saharsh Jul 27 '14 at 6:55
1
$\begingroup$

Observe that $\dfrac{\sqrt3+i}2=-w$ where $w$ is an imaginary cube root of unity

and consequently, $\dfrac{\sqrt3-i}2=-w^2$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How though? I don't understand. $\endgroup$ – Gummy bears Jul 27 '14 at 6:07
  • $\begingroup$ @Gummybears, See $n=3$ case here(en.wikipedia.org/wiki/Root_of_unity#Examples) $\endgroup$ – lab bhattacharjee Jul 27 '14 at 6:20
  • $\begingroup$ There has to be a simpler method? I haven't learnt this in class yet. $\endgroup$ – Gummy bears Jul 27 '14 at 6:25
  • $\begingroup$ @Gummybears believe me this one is the simplest method of all. A little bit of study of cube root of unity over internet gonna help you a lot in this. $\endgroup$ – Saharsh Jul 27 '14 at 6:54
  • $\begingroup$ Yes but not sure how the teachers would respond. $\endgroup$ – Gummy bears Jul 27 '14 at 7:02
1
$\begingroup$

Hint:

Supposing that one doesn't wish to make use of Euler's formula, we can achieve the result with binomial expansion.

$$\begin{align} z&=\left(\frac{\sqrt3}{2} + \frac{i}{2}\right)^{107} + \left(\frac{\sqrt3}{2} - \frac{i}{2}\right)^{107} \\&=\sum_{k=0}^{107}\binom{107}{k}\left(\frac{\sqrt3}{2}\right)^{107-k}\left[\left(\frac{i}{2}\right)^{k}+\left(\frac{-i}{2}\right)^{k}\right] \end{align}$$

Now what happens when $k$ is even? What about when $k$ is odd?

Notice that you can generalize the result for the situation $a^n+(\bar{a})^n$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Some facts to help you: $$\overline{x+y}=\overline x+\overline y$$$$\overline{xy}=\overline x\overline y$$$$\overline{x^y}=\overline x^{\overline y}$$ Similarly for subtraction, division, and roots. In general, conjugation distributes over nearly everything. One way to think about it: the $+$ sign and the $\times$ sign (and exponentiation) don't "know the difference" between $i$ and $-i$, so they don't care when you switch them around. (Complex numbers aren't mentioned when you define addition, etc.)

In fact, most functions are like this. For example, there is a way to define $\sin(z)$ even when $z$ is complex, and: $$\overline{\sin(z)}=\sin(\overline z)$$

The last fact to help you is that $a+\overline a$ is always real.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Well I knew the first 2. Didn't know about the last one. But the conjugate of 107 will still remain 107. I can't add two terms in powers and just ignore the power. $\endgroup$ – Gummy bears Jul 27 '14 at 7:11
  • $\begingroup$ The third fact - about exponentiation - tells you that your two terms are conjugates. Do you see why that is, or do you need more help? (The very last fact tells you that adding two conjugates gets you a real.) $\endgroup$ – Akiva Weinberger Jul 27 '14 at 7:15
  • $\begingroup$ What about the power? I understand that adding two conjugates, without power, will get you a real number. However what about the power? $\endgroup$ – Gummy bears Jul 27 '14 at 7:18
  • $\begingroup$ When I said "the two terms," I included the power. Call the comp. numbers you had $w$ and $\overline w$ (because I don't feel like writing it out - I am on a mobile.) What I'm saying is, $w^{107}$ is the conjugate of $\overline w^{107}$, by my third equation. (Remember that $\overline{107}=107$.) $\endgroup$ – Akiva Weinberger Jul 27 '14 at 7:20
  • $\begingroup$ Yep but then how to add the term and conjugate? $\endgroup$ – Gummy bears Jul 27 '14 at 7:54
0
$\begingroup$

Hint: ${\rm Im}(z)=(z-\overline z)/2i$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

You have the following:

$$ \left(a + ib\right)^n + \left(a - ib\right)^n $$

So which parts contribute to the imaginary part? The odd parts, right? You have:

\begin{align} \left(a + ib\right)^n + \left(a - ib\right)^n =& \sum_0^n\left(\binom{n}{k}a^k\left(ib\right)^{n - k}\right) + \sum_0^n\left(\binom{n}{k}a^k\left(-kb\right)^{n - k}\right) \\ =& \sum_0^n\left(\binom{n}{k}a^k\left(\left(ib\right)^{n - k} + \left(-ib\right)^{n - k}\right)\right) \end{align}

Regardless of imaginary values, an exponent of odd $(n - k)$ will result in $(ib)^{n -k} - (ib)^{n - k} = 0$. Furthermore, an even value of $2j = (n - k)$ results in $i^{2j} = \left(i^2\right)^j = \left(-1\right)^j=$ a real value! So any non-odd value of $(n - k)$ is irrelevant to computing the imaginary part.

In exponent form, what you have is:

\begin{align} a + ib =& r\cos\left(\theta\right) + ri\sin\left(\theta\right) = re^{i\theta} \\ a - ib =&r\cos\left(\theta\right) - ri\sin\left(\theta\right) = re^{-i\theta} \\ \left(a + ib\right)^n + \left(a - ib\right)^n =& r^ne^{in\theta} + r^ne^{-in\theta}\\ =& r^n\left(\cos\left(n\theta\right) + i\sin\left(n\theta\right) + \cos\left(n\theta\right) - i\sin\left(n\theta\right)\right)\\ =& 2r^n\cos\left(\pm n\theta\right) \end{align}

...which has no imaginary part!

EDIT:

Since you can gleen from the initial problem that:

\begin{align} \frac{\sqrt{3}}{2} + \frac{i}{2} =& \cos\left(60^\circ\right) + i\sin\left(60^\circ\right) = e^{i60^\circ} \\ \frac{\sqrt{3}}{2} - \frac{i}{2} =& \cos\left(-60^\circ\right) + i\sin\left(-60^\circ\right) = e^{-i60^\circ} \end{align}

We can conclude with $\theta = 60^\circ$ that the final result is:

\begin{align} 2\cdot1^{107}\cdot\cos\left(107\cdot60\right) =& 2\cos\left(17\cdot360^\circ + 300^\circ\right) \\ =& 2\cdot\cos\left(300^\circ\right) \\ =& 2\cdot\cos\left(-60^\circ\right) = 2\cdot\cos\left(60^\circ \right) \\ =& 2 \cdot\frac{\sqrt{3}}{2} = \sqrt{3} \end{align}

or

\begin{align} 2\cdot1^{107}\cdot\cos\left(107\cdot-60\right) =& 2\cos\left(-18\cdot360^\circ +60^\circ\right) \\ =& 2\cdot\cos\left(60^\circ\right) \\ =& 2 \cdot\frac{\sqrt{3}}{2} = \sqrt{3} \end{align}

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

EDIT: This post is false. Sorry.

In general: $$f(a+bi)+f(a-bi)$$always has a zero imaginary part, no matter what $f$ is. Reason: The two terms are conjugates of each other.* Adding two conjugates to each other gives you a real number.**

*This is true, because conjugation distributes over everything: $$\overline{f(z)}=f(\overline z)$$Basically, what conjugation does is it replaces $i$ with $-i$. If you think about it, there's no difference between the two! They both square to $-1$, and they both aren't real. ($-i$ isn't negative, because imaginary numbers don't have a concept of "being less than zero" - there is no concept of "less than" at all!)

**$(a+bi)+(a-bi)=2a$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't see how this is possible. Certainly it's true of the identity function, but I seriously doubt it's true of all functions. $\endgroup$ – Jared Jul 27 '14 at 6:48
  • 1
    $\begingroup$ It's true as long as $f$ is holomorphic, and is real for real inputs. For a counterexample to the general statement, take $f(z)=i$. $\endgroup$ – Peter Woolfitt Jul 27 '14 at 6:52
  • $\begingroup$ Oh, yes, I'm not sure how I managed to miss that obvious example. $\endgroup$ – Akiva Weinberger Jul 27 '14 at 6:58
  • $\begingroup$ How would I prove it. $\endgroup$ – Gummy bears Jul 27 '14 at 7:01
  • $\begingroup$ But what to do about the power? I can't add them if there is a power! Or can I.... $\endgroup$ – Gummy bears Jul 27 '14 at 7:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.