# Thread: complex numbers

1. ## complex numbers

Can someone please tell me why this statement is true:

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

Here's what I did:

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

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

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

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

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

powers of $i$ are periodic modulo 4.

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

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

Tonio

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

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