Let $$z=a+ib=rei$$θ where a,b,θ∈R and $$i=$$−$$1$$ then $$r=a2+b2=$$|z|and θ$$=$$$$tan$$−$$1$$⁡$$ba=arg$$⁡(z)now$$, |z|$$2=a2+b2=(a+ib)(a$$−$$ib)=zz$$¯⇒ $$1z=z$$¯|z|$$2and$$ $$z1z2z3$$….$$zn=z1z2z3$$…znif $$f(z)=1then$$ $$f(z) is called unimodular. In this $$casef(z)$$ can always be expressed as $$f(z)=ei$$α,α∈RAlso, eiα$$+ei$$β$$=ei$$α$$+$$β$$22cos$$⁡α−β$$2$$ and eiα−eiβ$$=ei$$α$$+$$β$$22isin$$⁡α−β$$2$$ where α,β∈RIf $$z1=1$$,$$z2=2$$,$$z3=3$$ and $$z1+z2+z3=1$$ then $$9z1z2+4z3z1+z2z3$$ is equal toThe value of $$tan$$⁡i ln⁡a−$$iba+ib$$ (where$$ $$i=$$−$$1$$ $$) is equal toIf |z−$$3i$$|$$=3$$ (where$$ $$i=$$−$$1$$ $$) and arg⁡z∈(0$$,π$$/2) then $$cot$$⁡(arg$$⁡$$z)−$$6z$$ is equal to

Question

Let $$z=a+ib=rei$$θ where a,b,θ∈R and $$i=$$−$$1$$ then  $$r=a2+b2=$$|z|and  θ$$=$$$$tan$$−$$1$$⁡$$ba=arg$$⁡(z)now$$, |z|$$2=a2+b2=(a+ib)(a$$−$$ib)=zz$$¯⇒ $$1z=z$$¯|z|$$2and$$  $$z1z2z3$$….$$zn=z1z2z3$$…znif   $$f(z)=1then$$ $$f(z) is called unimodular. In this $$casef(z)$$ can always be expressed as $$f(z)=ei$$α,α∈RAlso, eiα$$+ei$$β$$=ei$$α$$+$$β$$22cos$$⁡α−β$$2$$  and eiα−eiβ$$=ei$$α$$+$$β$$22isin$$⁡α−β$$2$$ where  α,β∈RIf $$z1=1$$,$$z2=2$$,$$z3=3$$ and $$z1+z2+z3=1$$ then $$9z1z2+4z3z1+z2z3$$ is equal toThe value of $$tan$$⁡i ln⁡a−$$iba+ib$$  (where$$ $$i=$$−$$1$$ $$)  is equal toIf |z−$$3i$$|$$=3$$  (where$$ $$i=$$−$$1$$ $$) and arg⁡z∈(0$$,π$$/2) then  $$cot$$⁡(arg$$⁡$$z)−$$6z$$ is equal to

Infinity Learn · Accepted Answer

∵$$z1=1$$,$$z2=2z3=3$$∴$$z12=1$$,$$z22=4$$,$$z32=9$$⇒$$z1z$$¯$$1=1$$,$$z2z$$¯$$2=4$$,$$z3z$$¯$$3=9$$∴$$1z1=z$$¯$$1$$,$$4z2=z$$¯$$2$$,$$9z3=z$$¯$$3now$$,$$9z1z2+4z3z1+z2z3$$ $$=z1z2z39z3+4z2+1z1=z1z2z39z3+4z2+1z1$$ $$=z1z2z3z$$¯$$1+z$$¯$$2+z$$¯$$3$$ $$=z1z2z3z1+z2+z3=z1z2z3z1+z2+z3=1$$⋅$$2$$⋅$$3$$⋅$$1=6$$ ∵$$z1+z2+z3=1a$$−$$iba+ib=$$|a−ib||$$a+ib$$|$$=a2+b2a2+b2=1$$∴  Let a−$$iba+ib=ei$$α⇒(a$$−$$ib)2a2+b2=ei$$α⇒ $$a2$$−$$b2a2+b2$$−$$i2aba2+b2=ei$$α⇒ ln⁡a−$$iba+ib=i$$α∴$$tan$$⁡iln⁡a−$$iba+ib=$$$$tan$$⁡$$(i$$⋅iα$$)=$$$$tan$$⁡$$($$−α$$)=$$−$$tan$$⁡α$$=$$−$$sin$$⁡α$$cos$$⁡α$$=2aba2$$−$$b2$$∵|z−$$3i$$|$$=3$$∴ z−$$3i=3ei$$α⇒ $$z=3i+3ei$$α$$=3ei$$π$$/2+ei$$α$$=3ei$$$$π$$$$4+$$α$$2$$⋅$$2cos$$⁡π$$4$$−α$$2$$∴ arg⁡$$(z)=$$π$$4+$$α$$2now$$,$$cot$$⁡$$(arg$$⁡$$z)−$$6z=$$$$cot$$⁡π$$4+$$α$$2$$−$$66e$$$$π$$$$4+$$α$$2cos$$⁡π$$4$$−α$$2=$$$$cot$$⁡π$$4+$$α$$2$$−e−$$1$$$$π$$$$4+$$α$$2cos$$⁡π$$2$$−π$$4+$$α$$2=$$$$cot$$⁡π$$4+$$α$$2$$−$$cos$$⁡π$$4+$$α$$2$$−isin⁡π$$4+$$α$$2sin$$⁡π$$4+$$α$$2=$$$$cot$$⁡π$$4+$$α$$2$$−$$cot$$⁡π$$4+$$α$$2+i=i$$

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!