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???
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???
$\displaystyle (1+i)^2=2i\Longrightarrow (1+i)^{200}=\left((1+i)^2\right)^{100}=(2i)^{100}= 2^{100}i^{100}=2^{100}$ , 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
$\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} $