<prompt> Kwenye ndege changamano, msambamba unaoundwa na pointi 0, z, frac1z, na z frac1z ina eneo frac3535 37. Ikiwa sehemu halisi ya z ni chanya, acha d iwe thamani ndogo zaidi ya left z frac1z kulia. Kokotoa d2.  <response>: Hebu z r cos theta i sin theta. Kishafrac1z frac1r cos theta i sin theta frac1r cos -theta i sin -theta frac1r cos theta - i sin theta.Kwa fomula ya kamba ya kiatu, eneo la pembetatu linaloundwa na 0, z r. cos theta ir sin theta na frac1z frac1r cos theta - fracir sin theta nifrac12 kushoto r cos theta kushoto -frac1r sin theta kulia - r sin theta kushoto frac1r cos theta kulia kulia sin theta cos theta,kwa hivyo eneo la msambamba ni2 dhambi theta cos theta sin 2 theta.Hivyo, sin 2 theta frac3537.Tunataka kupata thamani ndogo iwezekanavyo yaanzaalignkushoto z frac1z kulia kushoto r cos theta ir sin theta frac1r cos theta - fracir sin theta kulia kushoto r cos theta frac1r cos theta i kushoto r sin theta - frac1r sin theta kulia kulia.endalignMraba wa ukubwa huu nianzaalignkushoto r cos theta frac1r cos theta kulia2 kushoto r sin theta - frac1r sin theta kulia 2 r2 cos2 theta 2 cos2 theta frac1r cos2 theta r2 sin2 theta - 2 sin2 theta frac1r2 sin2 theta r2 frac1r2 2 cos2 theta - sin2 theta r2 frac1r2 2 cos 2 theta.endalignNa AM-GM, r2 frac1r2 ge 2. Pia,cos2 2 theta 1 - sin2 2 theta 1 - kushoto frac3537 kulia2 frac1441369, kwa hivyo cos 2 theta pm frac1237.Ili kupunguza usemi ulio hapo juu, tunachukua cos 2 theta -frac1237, sod2 2 - 2 cdot frac1237 frac5037. 