-3

$$Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 2$$

$$Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$$

I got that $Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 1 \ne 2$

And, that $\Big(({\frac{1+i}{1-i})^5\Big)} = i $ , which means that $Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$

Can you guys confirm that it's true? Thanks in advance!

In this image I got to$(\frac{1−\sqrt3+i+i\sqrt3}{2})^4$ and then, I'm not sure how to continue.

Martin Sleziak
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Ilan Aizelman WS
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  • $r\mathrm{e}^{i\theta} = r\left[\cos\theta +i\sin\theta\right]$ and compare terms. – Chinny84 Apr 28 '14 at 15:23
  • Because no1 helped me, I opened a new one. – Ilan Aizelman WS Apr 28 '14 at 15:29
  • @Chinny84 I can't do that. you can't find the angle and you can't find the r – Ilan Aizelman WS Apr 28 '14 at 15:30
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    @IlanAizelmanWS, it is considered improper here to repost your question just because you didn't like the outcome the first time. You should edit your first question, and wait. – vadim123 Apr 28 '14 at 15:38
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    Oh no! Nobody posted an answer for two whole hours, so you posted the question again? How inconsiderate of us... :| – Asaf Karagila Apr 28 '14 at 15:38
  • Yes you can, you can use $r=0.5$ and you should be able to determine an expression for $\tan \theta$ In terms of the real and imaginary component of the term in the bracket. I apologise that I didn't explain that we are using only the terms in the bracket, for which we can then raise to the 4th power. – Chinny84 Apr 28 '14 at 15:40
  • For the question number 2 \begin{align} \text{Im}\left(\frac{1+i}{1-i}\right)^5&=\text{Im}\left(\frac{1+i}{1-i}\cdot \frac{1+i}{1+i}\right)^5\ &=\text{Im}\left(\frac{(1+i)^2}{1^2-i^2}\right)^5\ &=\text{Im}\left(\frac{1+2i+i^2}{2}\right)^5\ &=\text{Im}\left(\frac{2i}{2}\right)^5\ \end{align} – Anastasiya-Romanova 秀 Apr 28 '14 at 15:47
  • I managed to get to that @V-Moy, as you can see in the image that I've added in my post. but I couldn't get to 2 in the first question. – Ilan Aizelman WS Apr 28 '14 at 15:53
  • @AsafKaragila: to be fair, he posted the duplicate to meta and it was migrated here >8( I didn't know about the previous post before I came up for air after answering. – robjohn Apr 28 '14 at 15:58
  • It's easy Sir. I'll give you a hint: $$ 1+\sqrt{3}i=2e^{\Large\frac{\pi}{3}i} $$ and $$ 1-i=\sqrt{2}e^{-\Large\frac{\pi}{4}i} $$ It's not difficult to find the $\text{abs}(z)$ and $\arg(z)$. Check Wiki. I hope this helps you. (ô‿ô) – Anastasiya-Romanova 秀 Apr 28 '14 at 16:02
  • Mr. @robjohn, for me it's not fair since I can't post my answer. (╥﹏╥) – Anastasiya-Romanova 秀 Apr 28 '14 at 16:05
  • Still didn't get it. – Ilan Aizelman WS Apr 28 '14 at 16:06
  • How did u go from $ 1 + \sqrt3i =$ to $2e^/sqrt pai/3i$? – Ilan Aizelman WS Apr 28 '14 at 16:07
  • @IlanAizelmanWS Are you kidding me? How come you learn complex number without knowing that thing?? If $z=x+iy$, then it can also be written as: $$ z=re^{i\theta} $$ where $r=\text{abs}(z)=\sqrt{x^2+y^2}$ and $\theta=\arg(z)=\arctan\left(\cfrac{y}{x}\right)$. Check Wiki or you can google about complex number for more detail explanation. – Anastasiya-Romanova 秀 Apr 28 '14 at 16:11
  • @V-Moy: sorry about that. I answered before it was closed. – robjohn Apr 28 '14 at 16:15
  • Mr. @robjohn: It's okay Sir. It's not your fault, no need to say sorry to me. ヅ – Anastasiya-Romanova 秀 Apr 28 '14 at 16:24
  • @robjohn you should either move your answer (and other's comment) to the other question and hard delete this question or reopen this and hard delete the other question. – achille hui Apr 29 '14 at 06:44
  • @achillehui: I answered this question as how to compute $$\mathrm{Re}\left[, \left(\frac{1-\sqrt3+i+i\sqrt3}2\right)^{\large4} ,\right]$$ I had not seen the original question (the one showing no work) before I answered this one, but nonetheless, to me, this is a different question than the other question, and my answer would not make sense as an answer to the original question. – robjohn Apr 29 '14 at 09:07

1 Answers1

2

Hint: Brute force works: $$ \begin{align} &\frac1{16}\mathrm{Re}\left[((1-\sqrt3)+i(1+\sqrt3))^4\right]\tag{1}\\ &=\frac1{16}\left(\color{#C00000}{(1-\sqrt3)^4}-\color{#00A000}{6(1-\sqrt3)^2(1+\sqrt3)^2}+\color{#C00000}{(1+\sqrt3)^4}\right)\tag{2}\\ &=\frac1{16}\left(\color{#C00000}{2(1+6\cdot3+9)}-\color{#00A000}{6(-2)^2}\right)\tag{3}\\[6pt] &=2\tag{4} \end{align} $$ where we have used the binomial theorem a few times $$ (x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4\tag{5} $$ Explanation:
$(1)$: copy problem
$(2)$: expand using $(5)$ and drop the terms with an $i$
$(3)$: expand the red terms using $(5)$ and drop the odd exponents ($-\sqrt3$ will cancel $+\sqrt3$)
$(3)$: also use that $(1-\sqrt3)(1+\sqrt3)=1-3=-2$
$(4)$: arithmetic

robjohn
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