I have to calculate $(3 + 3i)^{829}$
What I did:
$$|3 + 3i| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9}=\sqrt{18}$$
The argument of $3 + 3i$ is $45º$.
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2Most of your equations don't make sense. First make sure you are really writing equations (it doesn't look like you really are) when you write something as an equation, and then format it so we can actually see the math notation correctly, using http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – David K Sep 16 '16 at 18:05
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Also try to be more specific about what you did. We don't know what you multiplied by $829$ or what math error you got when you tried it. – David K Sep 16 '16 at 18:07
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$(3 + 3i)${^8}{^2}{^9} what does it mean? Would you like $(3 + 3i)^{829}$ or $(((3 + 3i)^{8})^2)^9$? – Robert Z Sep 16 '16 at 18:18
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@RobertZ I edited it – Funereal Sep 16 '16 at 18:21
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Now the equations are readable and true, well done! I simplified one of the formats, hoping it is still what you meant. – David K Sep 16 '16 at 18:24
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But so far everything looks good with your calculations, so what was the difficulty? – David K Sep 16 '16 at 18:25
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Yes, but I dont know how to calculate this raised to 829 – Funereal Sep 16 '16 at 18:26
2 Answers
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Note that $3+3i=3(1+i)=3\sqrt{2}e^{i\pi/4}=\sqrt{18}e^{i\pi/4}$. Hence $$(3+3i)^{829}=18^{829/2}\cdot e^{i\pi/4\cdot 829}= 18^{829/2}\cdot e^{i\pi/4\cdot (103\cdot 8+5)}= 18^{829/2}\cdot e^{i5\pi/4}$$ because $829=103\cdot 8+5$ and $(e^{i\pi/4})^8=e^{8i\pi/4}=e^{2\pi i}=1$.
P.S. Take a look to this similar exercise: Polar form of Complex numbers
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Nice answer but I think im a bit lost, you use a theory to calculate it? – Funereal Sep 16 '16 at 18:00
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Do you know how to use the polar form? What happens when you multiply two complex numbers in polar form? – Robert Z Sep 16 '16 at 18:26
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I think something escapes me. I have not used the polar way to get 18 ? or I have to keep using it ? – Funereal Sep 16 '16 at 18:33
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When you take a $n$-power in polar form you have to raise the modulus to $n$ and multiply the argument by $n$. Moreover in the argument you can throw away the integer multiples of $2\pi$ (i.e. $360º$) – Robert Z Sep 16 '16 at 18:42
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The key is in the title: you need to write the number in polar form: $$x+\mathrm i y = r(\cos\theta+\mathrm i \sin\theta)$$ where $r$ is the modulus of $x+\mathrm i y$ and $\theta$ is the argument.
In your case, you have $x + \mathrm i y = 3 + 3\mathrm i$.
The modulus is given by $\left|3+3\mathrm i\right|=\sqrt{3^2+3^2}=3\sqrt{2}$.
Since $3+3\mathrm i$ lies on the line $y=x$, and is in the upper-right quadrant. The argument is given by $\arg(3+3\mathrm i)=\frac{1}{4}\pi$.
That means that $r=3\sqrt{2}$ and $\theta=\frac{1}{4}\pi$, and hence $$3+3\mathrm i = 3\sqrt{2}\left(\cos(\tfrac{1}{4}\pi)+\mathrm i \sin(\tfrac{1}{4}\pi)\right)$$
De Moivre's Theorem tells is that $(\cos \theta + \mathrm i \sin \theta)^n = \cos (n\theta) + \mathrm i \sin (n\theta)$.
It follows that \begin{eqnarray*} (3+3\mathrm i)^{829} &=& (3\sqrt 2)^{829}\left(\cos(\tfrac{829}{4}\pi)+\mathrm i \sin(\tfrac{829}{4}\pi)\right) \end{eqnarray*}
The sine and cosine functions are $2\pi$-periodic, i.e. they repeat every $2\pi$ and so $\sin(x+2\pi)\equiv \sin x$ just like $\cos(x+2\pi) \equiv \cos x$ for all real $x$. We can subtract multiples of $2\pi=\frac{8}{4}\pi$ from $\frac{829}{4}\pi$ without changing the values of sine and cosine.
Since $829 = 4 \times 206 + 5$ it follows that $\frac{829}{4} = 206 + \frac{5}{4}$ and hence $\frac{829}{4}\pi = 206\pi + \frac{5}{4}\pi$.
Given that sine and cosine are $2\pi$ periodic:
\begin{eqnarray*} \cos(\tfrac{829}{4}\pi) &=& \cos(206\pi + \tfrac{5}{4}\pi) \\ \\ &=& \cos(\tfrac{5}{4}\pi) \\ \\ \sin(\tfrac{829}{4}\pi) &=& \sin(206\pi + \tfrac{5}{4}\pi) \\ \\ &=& \sin(\tfrac{5}{4}\pi) \end{eqnarray*}
It follows that \begin{eqnarray*} (3\sqrt 2)^{829}\left(\cos(\tfrac{829}{4}\pi)+\mathrm i \sin(\tfrac{829}{4}\pi)\right) &=& (3\sqrt 2)^{829}\left(\cos(\tfrac{5}{4}\pi)+\mathrm i \sin(\tfrac{5}{4}\pi)\right) \\ \\ &=& (3\sqrt 2)^{829}\left(-\tfrac{1}{\sqrt 2}-\tfrac{1}{\sqrt 2}\,\mathrm i \right) \\ \\ &=& -3^{829}\times 2^{414} \times (1+\mathrm i) \end{eqnarray*}
This number is massive and shouldn't be written down.
Fly by Night
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Thank you for your detailed reply , You've helped me a lot. Is it advisable to put the result ? Or would it be more correct to leave it ? – Funereal Sep 16 '16 at 20:01
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If the question was to compute $(3+3\mathrm i)^{829}$ then that is the answer. Perhaps they wanted the argument? In that case, it would be the argument of $-(1+\mathrm i)$ which is $-\frac{3}{4}\pi$. – Fly by Night Sep 16 '16 at 21:12