Question

find all complex number which makes the following equation true |z+1|=1 and |z^2+1|=1

Answer 1

Step-by-step explanation:

Let’s have another look:

z=a+i⋅bz=a+i⋅b

|z+1|=|(a+1)+i⋅b|=a2+b2+2a+1−−−−−−−−−−−−−√|z+1|=|(a+1)+i⋅b|=a2+b2+2a+1

or, introducing

f(a,b)=a2+b2+2a+1,f(a,b)=a2+b2+2a+1,

|z+1|=f(a,b)−−−−−√|z+1|=f(a,b)

|z2+1|=|(a+i⋅b)2+1|=(a2−b2+1)2+4a2b2−−−−−−−−−−−−−−−−−√|z2+1|=|(a+i⋅b)2+1|=(a2−b2+1)2+4a2b2

or, introducing

g(a,b)=(a2−b2+1)2+4a2b2g(a,b)=(a2−b2+1)2+4a2b2

The exact solution is found by solving

f(a,b)=1f(a,b)=1

g(a,b)=1g(a,b)=1

Two points satisfy the given equations:

z=−1/2±i⋅3–√/2

z1=z22;z2=z12