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0, under the reciprocal map defined on the extendedcomplex plane is the circle|w-12k|=|12k|The vertical linex=kconsists of all pointsz=k+iysuch thatx,y∈Rwherek6=0. So we may writew=1k+iy=kk2+y2-yk2+y2iThus our real and imaginary parts ofw=f(u,v)areu=kk2+y2,v=-yk2+y2,y∈RBut observe thatv=-yuk, which impliesy=-vku(1)And so, upon substituting into our initial expression foru, we haveu=kk2+ (-vku)2It is here advantageous to us to simplify and complete the square as followsu=kk2+ (-vku)2k=uk2+ (-vku)2k=k2u+v2k2uuk=k2u2+v2k20=u2+v2-uk-12k2=u2-uk+-12k2+v214k2=u-12k2+v2Finally, by recognition, we have|w-12k|=|12k|Note that we never made the restriction in (1) thatu6=0. This is because here we are working in theextended complex plane.3
2.6Limits and ContinuityNote:In order to save time, the extreme pedantism and wordiness that is seen in the text has here been avoided.For instance, substitution to evaluate a limit may take one line rather than half a page.1.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluatelimz→2i(z2-z)Let us simply substitute inz=2ias follows:(2i)2-2i=4i2+2i=-4+2i3.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluatelimz→1-i(|z|2-i¯z)Again, let us try substitution|1-i|2-i(1-i) =2-i+i2=2+1-i=3-i5.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluatelimz→πi(ez)Again, we need only substitute.eiπ=-19.) Use Theorem 2.6.2 and the basic limits (15) and (16) to computelimz→2-i(z2-z)We try substitution as follows(2-i)2-2+i=4-4i+i2+i-2=4-2-1-3i=1-3i11.) Use Theorem 2.6.2 and the basic limits (15) and (16) to computelimz→eiπ/4(z+1z)Yet again, we need only substitution for the following limit. In particular,limz→eiπ/4(z+1z) =eiπ/4+e-iπ/4=cos(π/4) +isin(π/4) +cos(-π/4) +isin(-π/4) =√24
13.) Use Theorem 2.6.2 and the basic limits (15) and (16) to computelimz→-iz4-1z+iFor the first time thus far, substitution will not suffice, for, obviously, it yields an indeterminate form(note that L’Hopital’s rule applies only to the reals). However, if we simply recognize thatz4-1=(z2-1)(z2+1) = (z-1)(z+1)(z+i)(z-i), then we have thatlimz→-iz4-1z+i=limz→-i(z+1)(z-1)(z-i) = (-i+1)(-i-1)(-2i) =4i19.) Consider the limitlimz→0zz2a.) What value foes the limit approach aszapproaches 0 along the real axis?b.) What value does the limit approach aszapproaches along the imaginary axis?c.) Do the answers from (a) and (b) imply that the limit exists? Explain.d.) What value does the limit approach aszapproaches along the liney=x?e.) What can you say about the limit in general?a.) Along thexaxis we havez=x+0y=x, thus the limit islimx→0xx2=limx→01=1b.) Along the imaginary axis we havex=0, implying thatz=iy, thus our limit becomeslimy→0yi-yi2=limy→0(-1)2=1c.) No. The limit must be the same along any of the infinitely (uncountably) many paths in the complexplane for it to exist in general.