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Answer 1 · 2017-11-27T08:48:39

a. $z=(-1+sqrt(3)i)/4=-1/4+(sqrt(3)i)/4$

b. $=-(1+sqrt(3)i)/64$

a. $z=v+w$

$4z=4v-4w=1+2sqrt(3)i-(2+sqrt(3)i)=-1+sqrt(3)i$

$z=(-1+sqrt(3)i)/4=-1/4+(sqrt(3)i)/4$

b. $(a+bi)^n=[((n),(0))b^0a^n-((n),(2))b^2a^(n-2)+((n),(2))b^4a^(n-4)-cdots]+i[((n),(1))b^1a^(n-1)-((n),(3))b^3a^(n-3)+((n),(5))b^5a^(n-5)-cdots]$

$(-1/4+(sqrt(3)i)/4)^5=[((5),(0))a^5-((5),(2))b^2a^3+((5),(4))b^4a]+i[((5),(1))ba^4-((5),(3))b^3a^2+((5),(5))b^5]$

$=[a^5-10b^2a^3+5b^4a]+i[5ba^4-10b^3a^2+b^5]$

$=[(-1/4)^5-10(sqrt(3)/4)^2(-1/4)^3+5(sqrt(3)/4)^4(-1/4)]+i[5(sqrt(3)/4)(-1/4)^4-10(sqrt(3)/4)^3(-1/4)^2+(sqrt(3)/4)^5]$

$=-1/64+i((-sqrt(3))/64)$

$=-1/64+(-sqrt(3)i)/64$

$=(-1-sqrt(3)i)/64$

$=-(1+sqrt(3)i)/64$