|z+1| = |z-1|
Im (z^2) = 2
Re z > -1
Re (1/z) < 1
Please just tell me what the graphs of these functions are and how would I know by looking at a function and telling how the graph is going to be
like for instance 0 < |z-1| < 1 is an annulus so what about the functions above
you must find the locus that each equation represents .
I will give you an example : Im(z^2) =2....do you understand what does it mean? it simply says that the imaginary part of an unknown complex number z^2 = (x+yi)^2
is equal to 2 . You must therefore find this imaginary part and put it equal to 2....lets do it. the z^2 =( x+yi)^2 yields after binomial aepansion that
z^2 = (x^2-y^2 )+(2xy)i. therefore the Im(Z^2) is 2xy and this must be equal to 2
therefore we have 2xy=2 or xy=1
now as you probably understand you must plot the function y =1/x this is of course a hyperbola known as reciprocal function...
check here Graphs of Reciprocal Functions (with worked solutions & videos)
to see its plot .
good luck with the others
If z= x+ iy then . . Graph that.
If z= x+ iy then Re(z)= x> -1
If z= x+ iy then so that
The best way to graph an inequality is to first graph the associated equality which is the boundary of the graph of the inequality.