If a and b are real numbers such that (2+)^4=a+b, where =-1+i√32, then a+b is equal to:

Question

If $a$ and $b$ are real numbers such that $(2+\alpha)^4=a+b\alpha$, where $\alpha=\dfrac{-1+i\sqrt3}2$, then $a+b$ is equal to:

Accepted Answer

$$\alpha  = \omega $$ as given $$\alpha  = {{ - 1 + i\sqrt 3 } \over 2}$$

$$ \Rightarrow {(2 + \omega )^4} = a + b\omega \,({\omega ^3} = 1)$$

$$ \Rightarrow {2^4} + {4.2^3}\omega  + {6.2^2}{\omega ^3} + 4.2.\,{\omega ^3} + {\omega ^4} = a + b\omega $$

$$ \Rightarrow 16 + 32\omega  + 24{\omega ^2} + 8 + \omega  = a + b\omega $$

$$ \Rightarrow 24 + 24{\omega ^2} + 33\omega  = a + b\omega $$

$$ \Rightarrow  - 24\omega  + 33\omega  = a + b\omega $$

$$ \Rightarrow a = 0,\,b = 9$$

Solutions · 1