Complex numbers and their imaginary parts

Question

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.

Answer 1

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

Answer 2

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

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

Answer 3

\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 $\Im (z)=0$

Answer 4

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

Answer 5

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}

Answer 6

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 $a^n+(\bar{a})^n$.

Answer 7

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!)

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

Answer 8

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.