• Mar 27th 2013, 02:54 AM
szak1592
|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
• Mar 27th 2013, 03:02 AM
Plato
Quote:

Originally Posted by szak1592
|z+1| = |z-1|

The above is the set of all points which are equally distance from 1 to -1.
• Mar 27th 2013, 03:05 AM
szak1592
thanks and what about the others
• Mar 27th 2013, 03:12 AM
MINOANMAN
Szak1952

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
MINOAS
• Mar 27th 2013, 05:52 AM
HallsofIvy
If z= x+ iy then $\displaystyle z^2= x^2- y^2+ 2xyi$. $\displaystyle Im(z^2)= 2xy= 2$. Graph that.
If z= x+ iy then $\displaystyle \frac{1}{z}= \frac{1}{x+ iy}\frac{x- iy}{x- iy}= \frac{x- iy}{x^2+ y^2}$ so that $\displaystyle Re(1/z)= \frac{x}{x^2+ y^2}< 1$