Question 2.6.4: Computing Limits with Theorem 2.2. Use Theorem 2.2 and the b......
Computing Limits with Theorem 2.2
Use Theorem 2.2 and the basic limits (15) and (16) to compute the limits
(a) \underset{z\rightarrow i}{lim}\frac{(3+i)z^{4}-z^{2}+2z}{z+1}
(b) \underset{z\rightarrow 1+\sqrt{3}i }{lim}\frac{z^{2}-2z+4}{z-1-\sqrt{3}i }
\underset{z\longrightarrow z_{0}}{lim} c=c, c a complex constant, (15)
\underset{z\longrightarrow z_{0}}{lim} z=z_{0} (16)
(a) By Theorem 2.2(iii) and (16), we have:
\underset{z\longrightarrow z_{0}}{lim} f(z). g(z)=L. M, Theorem 2.2 (iii)
\underset{z\longrightarrow i}{lim} z^{2}= \underset{z\longrightarrow i}{lim} z.z=\left(\underset{z\longrightarrow i}{lim} z\right) .\left(\underset{z\longrightarrow i}{lim} z\right)=i.i=-1Similarly,\underset{z\longrightarrow i}{lim}z^{4}=i^{4}=1. Using these limits, Theorems 2.2(i), 2.2(ii), and the limit in (16), we obtain:
\underset{z\longrightarrow z_{0}}{lim} (cf(z))=cL, \text{c a complex constant,} Theorems 2.2(i),
\underset{z\longrightarrow z_{0}}{lim} (f(z)\pm g(z))=L\pm M, Theorems 2.2(ii),
\underset{z\longrightarrow i}{lim} ((3+i)z^{4}-z^{2}+2z)= (3+i)\underset{z\longrightarrow i}{lim}z^{4} – \underset{z\longrightarrow i}{lim}z^{2} +2\underset{z\longrightarrow i}{lim}z\\=(3+i)(1)-(-1)+2(i)\\=4+3i,and \underset{z\longrightarrow i}{lim}(z+1)=1+i. Therefore, by Theorem 2.2(iv ), we have:
\underset{z\longrightarrow z_{0}}{lim} \frac{f(z)} {g(z)}=\frac{L}{M} , \text{provided} M\neq 0 Theorem 2.2(iv )
\underset{z\longrightarrow i}{lim}\frac{(3+i)z^{4}-z^{2}+2z}{z+1} = \frac{\underset{z\longrightarrow i}{lim}((3+i)z^{4}-z^{2}+2z)}{\underset{z\longrightarrow i}{lim}(z+1)}=\frac{4+3i}{1+i}After carrying out the division, we obtain \underset{z\longrightarrow i}{lim}\frac{(3+i)z^{4}-z^{2}+2z}{z+1} = \frac{7}{2}-\frac{1}{2}i.
(b) In order to find \underset{z\rightarrow 1+\sqrt{3}i }{lim}\frac{z^{2}-2z+4}{z-1-\sqrt{3}i } , we proceed as in (a):
\underset{z\longrightarrow1+\sqrt{3} i}{lim}(z^{2}-2z+4) = \left(1+\sqrt{3}i \right) ^{2}-2 \left(1+\sqrt{3}i \right)+4\\=-2+2\sqrt{3}i-2-2\sqrt{3}i+4=0,and \underset{z\longrightarrow1+\sqrt{3} i}{lim}(z-1-\sqrt{3}i) =1+\sqrt{3}i-1-\sqrt{3}i=0. It appears that we cannot apply Theorem 2.2(iv ) since the limit of the denominator is 0. However, in the previous calculation we found that 1 +\sqrt{3}i is a root of the quadratic polynomial z^{2}−2z+4. From Section 1.6, recall that if z_{1} is a root of a quadratic polynomial, then z−z_1 is a factor of the polynomial. Using long division, we find that
z^{2}-2z+4=\left(z-1+\sqrt{3}i\right) \left(z-1-\sqrt{3}i\right) .
See (5) in Section 1.6 and Problems 25 and 26 in Exercises 1.6. Because z is not allowed to take on the value 1 +\sqrt{3}i in the limit, we can cancel the common factor in the numerator and denominator of the rational function. That is,
az^{2} + bz + c = a(z − z_{1})(z − z_{2}). (5) in Section 1.6
\underset{z\rightarrow 1+\sqrt{3}i}{lim }\frac{z^{2} – 2z + 4}{z-1-\sqrt{3}i} =\underset{z\rightarrow 1+\sqrt{3}i}{lim }\frac{(z − 1+\sqrt{3}i)(z −1-\sqrt{3}i)}{(z − 1-\sqrt{3}i)}\\=\underset{z\rightarrow 1+\sqrt{3}i}{lim } (z-1+\sqrt{3}i) .By Theorem 2.2(ii) and the limits in (15) and (16), we then have
\underset{z\rightarrow 1+\sqrt{3}i}{lim } (z-1+\sqrt{3}i)=1+\sqrt3 i-1+\sqrt{3}i=2\sqrt{3}i.
Therefore, \underset{z\rightarrow 1+\sqrt{3}i}{lim } \frac{z^{2}-2z+4}{z-1-\sqrt{3}i } =2\sqrt{3} i.