# How do i find $(1+i)^{100}?$

How do I find $(1+i)^{100}$ without expanding $(1+i)$ 100 times?

Is there a quicker way to do this?

The hint was to find the modulus and argument of $1+i$ which I've got as $\sqrt{2}$ and $\pi/4$ but I'm not sure what to do from here.

$$(1+i)^2=2i\qquad \Longrightarrow\qquad (1+i)^4=-4.$$

• Nice solution, it reduces the calculation directly to integers. Commented Jun 26, 2014 at 20:01
• The simplest solution! Nice! Commented Jun 26, 2014 at 20:04
• No this solution distracts from the whole point of the question Commented Jun 26, 2014 at 22:17
• @JohnFernley Another barometer problem? Commented Jun 27, 2014 at 8:44
• Any difference from $(1+i)=\sqrt{2}e^{\frac{i\pi}{4}}$ in simplicity is very slight Commented Jun 27, 2014 at 13:47

Hint: $$(1+i)=\sqrt{2}e^{\frac{i\pi}{4}}$$

Hint:

$1+i=\sqrt{2} e^{i\pi/4}$

Therefore, $(1+i)^{100}= [\sqrt{2} e^{i\pi/4}]^{100}=\sqrt{2}^{100}e^{i100\pi/4}=[{{2^{0.5}}}]^{100}e^{i(24\pi+\pi)}=2^{50}e^{i\pi}=2^{50}(-1)=\boxed{-2^{50}}$

Your job: how did we get $1+i=\sqrt{2} e^{i\pi/4}$?

To do this kind of problem in general, look at the polar form of a complex number:

http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Forms.aspx

You can use De Moivre's formula, $1+i=\sqrt{2}(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4})$, so $(1+i)^{100}=(\sqrt{2})^{100}(\cos 25\pi+i \sin 25\pi)$

The ( at least a ) key is that the powers are periodic (argument-wise) .One way of doing Complex multiplication is by doing it geometrically, you multiply two numbers by multiplying their respective moduli and by adding their respective arguments. Notice that $$1+i$$ lies on the line $$y=x$$, so that its argument is $$\pi/4$$. What happens if you go around $$\frac {2k\pi}{\pi/4}$$ times? The key is that $$\pi/4$$ is "commensurate" to $$2\pi$$, meaning it divides exactly (an integer number of times) into $$2 \pi$$.