<s>[INST] Express $(\sqrt{3}+i)^{6-i}$ in the form $x+iy$ I have used that  $2e^{i\pi /6}=\sqrt{3}+i $ and I have reached that $(\sqrt{3}+i)^{6-i}= 2^{6-i}(-e^{\pi /6}) $ but I do not know how to pass $2^{6-i}$ to the form $x + iy$, could anyone help me? Thank you.
 [/INST] $$\left( \sqrt { 3 } +i \right) ^{ 6-i }=\left( 2{ e }^{ \frac { \pi  }{ 6 } i } \right) ^{ 6-i }={ 2 }^{ 6-i }{ e }^{ \pi i }{ e }^{ \frac { \pi  }{ 6 }  }=64{ e }^{ \frac { \pi  }{ 6 }  }\left( { e }^{ i\left( \pi -\ln { 2 }  \right)  } \right) =\\ =64{ e }^{ \frac { \pi  }{ 6 }  }\left( \cos { \left( \pi -\ln { 2 }  \right)  } +i\sin { \left( \pi -\ln { 2 }  \right)  }  \right) =\\ =64{ e }^{ \frac { \pi  }{ 6 }  }\left( -\cos { \left( \ln { 2 }  \right)  } +i\sin { \left( \ln { 2 }  \right)  }  \right) =\\ =-64{ e }^{ \frac { \pi  }{ 6 }  }\left( \cos { \left( -\ln { 2 }  \right)  } +i\sin { \left( -\ln { 2 }  \right)  }  \right) $$
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