Let z=a+ib=reiθ 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)=eiα,α∈RAlso, eiα+eiβ=eiα+β22cos⁡α−β2  and eiα−eiβ=eiα+β22isin⁡α−β2 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

∵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=eiα⇒(a−ib)2a2+b2=eiα⇒ a2−b2a2+b2−i2aba2+b2=eiα⇒ ln⁡a−iba+ib=iα∴tan⁡iln⁡a−iba+ib=tan⁡(i⋅iα)=tan⁡(−α)=−tan⁡α=−sin⁡αcos⁡α=2aba2−b2∵|z−3i|=3∴ z−3i=3eiα⇒ z=3i+3eiα=3eiπ/2+eiα=3eiπ4+α2⋅2cos⁡π4−α2∴ arg⁡(z)=π4+α2now,cot⁡(arg⁡z)−6z=cot⁡π4+α2−66eπ4+α2cos⁡π4−α2=cot⁡π4+α2−e−1π4+α2cos⁡π2−π4+α2=cot⁡π4+α2−cos⁡π4+α2−isin⁡π4+α2sin⁡π4+α2=cot⁡π4+α2−cot⁡π4+α2+i=i