First I wrote the complex number in polar form:
0,5+i0,5= r(cosv+isinv)
tan(v)=0,5/0,5 =1 v=45
r=√(0,5^2+0,5^2)=√0,5
--> 0,5+i0,5= √0,5(cos45+isin45)
Then I used de Moivres formula and got the following answer
(0,5+i0,5)^7= (√0,5)^7 (cos(45×7)+ isin(45×7)=
(√0,5)/8(cos315+isin315)= 0,0625+(-i0,0625)= 0,0625-i0,0625.
Does anyone think the answer should be diffrent?
1) Binomial theorem.
$(\frac 12 + \frac 12i)^7 = \frac 1{2^7}(1+i)^7 = \frac 1{128}(1 + 7i + \frac {7*6}2i^2 + \frac {7*6*5}{2*3} i^3 + \frac {7*6*5}{2*3}i^4 + \frac {7*6}{2} i^5+7i^6 + i^7)=$
$ \frac 1{128}(1 + 7i - 21 - 35i + 35 + 21i -7 -i)=\frac 1{128}(8-8i)=\frac 1{16}-\frac 1{16}i$
..... which was way too hard....
2) polar coordinates.
$\frac 12 + \frac 12 i = r(\cos \theta + i \sin \theta)$.
$r = \sqrt {(\frac 12)^2 + (\frac 12)^2} = \sqrt{\frac 12}=\frac {\sqrt 2}2=(\frac 12)^{\frac 12}$ and $\theta = \arctan \frac {\frac 12}{\frac 12}=\arctan 1 = \frac {\pi} 4$.
(You really should stop relying on decimals and you should start using radians.)
So $(\frac 12 + \frac 12i)^7 = (\frac {1}{2})^{\frac 72}(\cos \frac{7\pi}4 + i \sin \frac {7\pi}4)=$
$(\frac {1}{2})^{\frac 72}(\frac {\sqrt 2}2 - \frac {\sqrt 2}2i)=(\frac 12)^{\frac 72}((\frac 12)^{\frac 12} - i (\frac 12)^{\frac 12})= \frac 1{2^4}(1-i)=$
$\frac 1{16} - \frac 1{16}i$.