1. ## complex numbers

Can someone please tell me why this statement is true:

$\displaystyle (1+i)^{200} = 2^{100}$

Here's what I did:

$\displaystyle ((1+i)^2)^{100} = 2^{100} \rightarrow (1+i)^2 = 2 \rightarrow 2i \neq 2$

as in it's false... so what's my mistake???

2. because $\displaystyle i^{100} = (i^4)^{25} = 1^{25} = 1$

The equation $\displaystyle x^{100} = 2^{100}$ doesn't mean $\displaystyle x = 2$ is the only solution , there are other solutions such as $\displaystyle x = -2 , 2i , -2i$ etc .

3. $\displaystyle (1+i)^2=2i\Longrightarrow (1+i)^{200}=\left((1+i)^2\right)^{100}=(2i)^{100}= 2^{100}i^{100}=2^{100}$ , since the

powers of $\displaystyle i$ are periodic modulo 4.

And your mistake was in $\displaystyle a^{100}=b^{100}\Longrightarrow a=b$, since it should be $\displaystyle a^{100}=b^{100}\Longrightarrow \left(\frac{a}{b}\right)^{100}=1\Longrightarrow \frac{a}{b}$ is

one of the 100 different roots of unit of order 100...or, of course, $\displaystyle a=b=0$

Tonio

4. $\displaystyle \left( {1 + i} \right)^{200} = \left( {\sqrt 2 } \right)^{200} \left[ {\cos \left( {\frac{\pi } {4}} \right) + i\sin \left( {\frac{\pi } {4}} \right)} \right]^{200}$
$\displaystyle =\left( {\sqrt 2 } \right)^{200} \left[ {\cos \left( {\frac{{200\pi }} {4}} \right) + i\sin \left( {\frac{{200\pi }} {4}} \right)} \right] = 2^{100}$