How to solve $\sqrt{2}\cdot(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)^{71}$ by hand?
Unfortunately I have no idea how to work through that task. Is there any technique for that?
My solution book says that $1+i$ is the correct solution.
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1Use polar form. – Apr 02 '17 at 14:42
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Do you know De Moivre's Theorem? – Edward Evans Apr 02 '17 at 14:42
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2Try $(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)^n$ for $n=0,1,2,3,4$ – Henry Apr 02 '17 at 14:42
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@ÍgjøgnumMeg, no I do not know De Moivre's Theorem – jublikon Apr 02 '17 at 14:44
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@jublikon Then the only way is binomial theorem. – Apr 02 '17 at 14:53
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1@A---B: Augh, no! Exponentiation by squaring is much cleaner here. – hmakholm left over Monica Apr 02 '17 at 14:59
6 Answers
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Even directly:
$$\sqrt2\left(\frac1{\sqrt2}-\frac1{\sqrt2}i\right)^{71}=\frac1{2^{35}}\left(1-i\right)^{71}$$
and now:
$$(1-i)^2=-2i\implies(1-i)^4=(-2i)^2=-4\implies(1-i)^{68}=\left[(1-i)^4\right]^{17}=-4^{17}\implies$$
$$(1-i)^{71}=-4^{17}(-2i)(1-i)=-4^{17}(-2-2i)=2^{35}+2^{35}i$$
DonAntonio
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HINT:
Using this, $$\dfrac1{\sqrt2}-\dfrac i{\sqrt2}=\cos\left(-\dfrac\pi4\right)+i\sin\left(-\dfrac\pi4\right)=e^{-i\pi/4}$$ by How to prove Euler's formula: $e^{it}=\cos t +i\sin t$?
lab bhattacharjee
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One way of going about this would be to convert to polar coordinates: recall that we have $re^{i \theta} = r(\cos(\theta) + i \sin(\theta))$. Igoring the factored-out $\sqrt{2}$ for now, you can check that we have $r=1$ ($r$ being the modulus of the complex number inside the parentheses). Therefore, we just need to find a $\theta$ such that $\cos(\theta) = 1/\sqrt{2}$ and $\sin(\theta) = -1/\sqrt{2}$.
After you've made the conversion, notice how much easier it is to compute powers of $e^{i \theta}$. When you're finished exponentiating, you can convert back to rectangular coordinates if desired, multiplying the result by the previously-ignored $\sqrt{2}$ to finish up.
Kaj Hansen
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It doesn't seem to be that much of a mess to me. Square the base three times and you end at $1$ ... – hmakholm left over Monica Apr 02 '17 at 14:59
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You're right @HenningMakholm. I guess what I mean to say is that, in general, exponentiating things of the form $(x + iy)$ has a high likelihood of being a mess. I admit I didn't consider this specific case first because of that. – Kaj Hansen Apr 02 '17 at 15:07
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Without De Moivre or polar coordinates: Note that by direct computation, $$\Bigl(\tfrac1{\sqrt2}-\tfrac1{\sqrt2}i\Bigr)^2 = -i $$ so $$\left(\tfrac1{\sqrt2}-\tfrac1{\sqrt2}i\right)^{71} = \left(\tfrac1{\sqrt2}-\tfrac1{\sqrt2}i\right)(-i)^{35} $$ and then $(-i)^2 = -1$ so $$ (-i)^{35} = -i\cdot (-1)^{17} = i $$
(This is just what you get from carrying out exponentiation by squaring by hand).
hmakholm left over Monica
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The easy way is to convert the complex number in brackets in its polar form: $$(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i) = re^{i \theta}$$ where $\sqrt{ (\frac{1}{\sqrt{2}})^2 + (- \frac{1}{\sqrt{2}})^2} = 1$ and $\theta = 7 \pi /4$, i.e on the line $f(x) = -x$. Thus the original expression becomes $$\sqrt{2}(e^{i 7 \theta / 4})^{71} $$ or $$\sqrt{2}(e^{i 497 \theta / 4}) $$ Then you can simply compute $e^{i 497 \theta /4} = \cos(497 \theta /4) + i \sin(497 \theta /4)$ and get your answer.
Edit: if you want a final answer by hand you can break down the exponent 497/4 so that it's easier to compute the sine and cosine.
