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Algebra of complex numbers

Question

Let z=a+ib=re where a,b,θR and i=1 

then  r=a2+b2=|z|

and  θ=tan1ba=arg(z)

now, |z|2=a2+b2=(a+ib)(aib)=zz¯

 1z=z¯|z|2

and  z1z2z3.zn=z1z2z3zn

if   f(z)=1then f(z) is called unimodular. In this case
f(z) can always be expressed as f(z)=e,αR

Also, e+e=eiα+β22cosαβ2  and 

ee=eiα+β22isinαβ2 where  α,βR

Moderate
Question

If z1=1,z2=2,z3=3 and z1+z2+z3=1 then 9z1z2+4z3z1+z2z3 is equal to

Solution

z1=1,z2=2z3=3z12=1,z22=4,z32=9z1z¯1=1,z2z¯2=4,z3z¯3=91z1=z¯1,4z2=z¯2,9z3=z¯3

now,

9z1z2+4z3z1+z2z3 =z1z2z39z3+4z2+1z1=z1z2z39z3+4z2+1z1 =z1z2z3z¯1+z¯2+z¯3 =z1z2z3z1+z2+z3=z1z2z3z1+z2+z3=1231=6 z1+z2+z3=1

Question

The value of tani lnaiba+ib  (where i=1 )  is equal to

Solution

aiba+ib=|aib||a+ib|=a2+b2a2+b2=1  Let aiba+ib=e(aib)2a2+b2=e a2b2a2+b2i2aba2+b2=e lnaiba+ib=tanilnaiba+ib=tan(i)=tan(α)=tanα=sinαcosα=2aba2b2

Question

If |z3i|=3  (where i=1 ) and argz(0,π/2) then  cot(argz)6z is equal to

Solution

|z3i|=3 z3i=3e z=3i+3e=3e/2+e=3eiπ4+α22cosπ4α2 arg(z)=π4+α2

now,

cot(argz)6z=cotπ4+α266eπ4+α2cosπ4α2=cotπ4+α2e1π4+α2cosπ2π4+α2=cotπ4+α2cosπ4+α2isinπ4+α2sinπ4+α2=cotπ4+α2cotπ4+α2+i=i



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