Question 1066912
A) The greatest and least values of {{{abs(z[1]+z[2])}}}
will happen when the {{{theta}}} angles for the polar form of {{{z[1]}}} and {{{z[2]}}}
are the same or {{{180^o}}} apart.
When that happens,
{{{abs(z[1]+z[2])=abs(abs(z[1])+abs(z[2]))}}} or
{{{abs(z[1]+z[2])=abs(abs(z[1])-abs(z[2]))}}}
 
In this case:
{{{abs(z[2])=sqrt(3^2+4^2)=5}}} ,
so the greatest and least values asked for are
{{{"max("}}}{{{abs(z[1]+z[2])}}}{{{")"=6+5=highlight(11)}}} and
{{{"min("}}}{{{abs(z[1]+z[2])}}}{{{")"=6-5=highlight(1)}}}
 
B) Let it be
{{{z=a+ib}}} , where {{{a}}} and {{{b}}} are real numbers.
{{{Re(z)=Re(a+ib)=a}}}
{{{z-(2+i)=(a+ib)-(2+i)=a+ib-2-i=a-2+ib-i=(a-2)+i(b-1)}}}
{{{abs(z-(2+i))=abs((a-2)+i(b-1))=sqrt((a-2)^2+(b-1)^2)}}}
{{{abs(z-(2+i))<=1}}} means {{{sqrt((a-2)^2+(b-1)^2)<=1}}}
Squaring both sides of the equal sign in that real numbers equation
gives us the real numbers equation
{{{(a-2)^2+(b-1)^2<=1}}} ,
which contains all the solutions for {{{sqrt((a-2)^2+(b-1)^2)<=1}}} ,
and could have new, extraneous solutions only if {{{(a-2)^2+(b-1)^2<0}}} .
Since {{{(a-2)^2+(b-1)^2>=0}}} for real numbers,
there will be no extraneous solutions.
All we have to do is solve {{{(a-2)^2+(b-1)^2<=1}}} for {{{a}}} .