解答
化简 (k+q)2e−iqa+(k−q)2eiqa
解答
(2q2cos(aq)+2k2cos(aq))−4ikqsin(aq)
求解步骤
(k+q)2e−iqa+(k−q)2eiqa
e−iqa=cos(aq)−isin(aq)
=(q+k)2(cos(aq)−isin(aq))+eiaq(−q+k)2
eiqa=cos(aq)+isin(aq)
=(q+k)2(cos(aq)−isin(aq))+(−q+k)2(cos(aq)+isin(aq))
=(k+q)2(cos(aq)−isin(aq))+(k−q)2(cos(aq)+isin(aq))
(k+q)2:k2+2kq+q2
=(k2+2kq+q2)(cos(qa)−isin(qa))+(k−q)2(cos(qa)+sin(qa)i)
(k−q)2:k2−2kq+q2
=(k2+2kq+q2)(cos(qa)−isin(qa))+(k2−2kq+q2)(cos(qa)+sin(qa)i)
乘开 (k2+2kq+q2)(cos(qa)−isin(qa)):k2cos(aq)−ik2sin(aq)+2kqcos(aq)−2ikqsin(aq)+q2cos(aq)−iq2sin(aq)
=k2cos(aq)−ik2sin(aq)+2kqcos(aq)−2ikqsin(aq)+q2cos(aq)−iq2sin(aq)+(k2−2kq+q2)(cos(qa)+sin(qa)i)
乘开 (k2−2kq+q2)(cos(qa)+sin(qa)i):k2cos(aq)+ik2sin(aq)−2kqcos(aq)−2ikqsin(aq)+q2cos(aq)+iq2sin(aq)
=k2cos(aq)−ik2sin(aq)+2kqcos(aq)−2ikqsin(aq)+q2cos(aq)−iq2sin(aq)+k2cos(aq)+ik2sin(aq)−2kqcos(aq)−2ikqsin(aq)+q2cos(aq)+iq2sin(aq)
化简 k2cos(aq)−ik2sin(aq)+2kqcos(aq)−2ikqsin(aq)+q2cos(aq)−iq2sin(aq)+k2cos(aq)+ik2sin(aq)−2kqcos(aq)−2ikqsin(aq)+q2cos(aq)+iq2sin(aq):2q2cos(aq)−4ikqsin(aq)+2k2cos(aq)
=2q2cos(aq)−4ikqsin(aq)+2k2cos(aq)
将 2q2cos(aq)−4ikqsin(aq)+2k2cos(aq) 改写成标准复数形式:(2q2cos(aq)+2k2cos(aq))−4kqsin(aq)i
=(2q2cos(aq)+2k2cos(aq))−4kqsin(aq)i