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I saw a pattern while evaluating some other powers of similar complex number so I tried to calculate the above question by expanding it, please tell me if it is correct...?

$(1+i)^{2} = 2i$

$(1+i)^{4}$ = $(2i)^2$ = $-4$

$(1+i)^{8} = (-4)^2 = 16$

similarly,

$(1+i)^{5404}$ = $(1+i)^{4096}$ $(1+i)^{256}$ $(1+i)^{32}$ $(1+i)^{16}$ $(1+i)^{4}$

$(1+i)^{5404}$ = $(256)^{256} (256)^{16} (256)^2 (16)^2 (-4)$

But $(256)^{256} = \infty$

Now how can I solve it? Is my method wrong?

Nathaniel B
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shivani
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    Your second line is enough to ovtain the answer. Note that $5404/4=1351$. – André Nicolas Jul 03 '16 at 04:23
  • Hint: $1+i=\sqrt2\exp(i\pi/4)$. – Clayton Jul 03 '16 at 04:25
  • But again $(-4)^{1351}=infinity !$ @AndréNicolas – shivani Jul 03 '16 at 04:30
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    As a typesetting note, use brackets to make sure everything you intend to be raised is actually raised. Otherwise, only the first character will be put in the superscript. compare $a^bcde$ $a^bcde$ to $a^{bcde}$ $a^{bcde}$ – JMoravitz Jul 03 '16 at 04:30
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    @shivani Why do you believe $(-4)^{1351}=\infty$? Surely it must be a finite quantity, albeit a large one. – zz20s Jul 03 '16 at 04:31
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    You should get $-(4^{1351})$. Large negative, but not infinite. However, your calculator may get upset if you ask it to compute this. – André Nicolas Jul 03 '16 at 04:32
  • Because when I wrote it in calculator, it showed infinity...!! – shivani Jul 03 '16 at 04:34
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    Perhaps you think $256^{256}=\infty$ because you are trying to plug this into a calculator. This number will be larger than most pocket calculators can handle as it will have $617$ digits, but it is of course a finite number. It is just larger than the largest number that it can think of. Calculators are not the end-all be-all of computation. There comes times where doing math by hand is better than doing math by a calculator since they can be wrong. – JMoravitz Jul 03 '16 at 04:35
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    Simple solution, stay away from the calculator, something that is surprisingly often useful to do. – André Nicolas Jul 03 '16 at 04:35
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    @shivani i hope you've learned to not EVER trust a calculator. Not ever does it make sense that finite to finite power is infinite. It is illogical – Zelos Malum Jul 03 '16 at 05:10
  • yes @ZelosMalum its really true... finite to finite power should always be finite and we 'll get a finite number no matter how large it is... I will never forget that... – shivani Jul 03 '16 at 05:13
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    Good, then all is learned and well :-) – Zelos Malum Jul 03 '16 at 05:24

3 Answers3

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Use Euler's Formula $$e^{i\theta}=\cos (\theta) + i\sin (\theta)$$ since $1+i=\sqrt{2} (\cos\frac{\pi}{4} +i\sin\frac{\pi}{4})= \sqrt{2} e^{i\frac{\pi}{4}}$, so $(1+i)^{5404} =(\sqrt{2} e^{i\frac{\pi}{4}})^{5404}= 2^{2702} e^{i\cdot 1351\pi}= 2^{2702} e^{i\pi}= -2^{2702}$.

Victor Chen
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Here is a general method without using Euler's formula, and formulas for this expansion for any $k$ exponent.

Since we can view multiplication of $i$ as a rotation by 90 degrees, which is periodic with return to the point of origin after $4$ rotations, we see that $i$ is an element of order 4 (if you've seen a little algebra).

This gives us:

$i^{n}=i$ whenever $n\cong 1 (\mod 4)$

$i^{n}=-1$ whenever $n\cong 2 (\mod 4)$

$i^{n}=-i$ whenever $n\cong 3 (\mod 4)$

$i^{n}=1$ whenever $n\cong 0 (\mod 4)$

Breaking up by these cases, and searching for a pattern from a few computations you can conjecture the following formulas:

If $k\cong 1 (\mod 4)$, we have: \begin{equation*} (1+i)^{k}=(-4)^{\lfloor{k/4}\rfloor}(1+i) \end{equation*} if $k\cong 2 (\mod 4)$, we have: \begin{equation*} (1+i)^{k}=(-4)^{\lfloor{k/4}\rfloor}2i \end{equation*} if $k\cong 3 (\mod 4)$, we have: \begin{equation*} (1+i)^k=(-4)^{\lfloor{k/4}\rfloor}(2i-2) \end{equation*} if $k\cong 0 (\mod 4)$, we have: \begin{equation*} (1+i)^k=(-4)^{k/4} \end{equation*}

All of which can be proved by induction without too much trouble.

operatorerror
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Thanks everyone... Here is the solution as I actually wanted to calculate it. But the main problem was I used calculator instead of my pen!

$(1+i)^{4} = -4$ &
$(1351)(4) = 5404$

$(1+i)^{5404} = (-4)^{1351}$

$(-4)^{1351} = (-4)^{1350} (-4)$

$(-4)^{1351} = (4)^{675} (4)^{675} (-4)$

$(-4)^{1351} \approx (-2.2) (10)^{812}$

$(1+i)^{5404} \approx (-2.2) (10)^{812}$

shivani
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