Question 2.6.2: An Epsilon-Delta Proof of a Limit. Prove that limz→1+i (2 + ......

According to Definition 2.8, \underset{z\rightarrow 1+i}{lim} (2+i)z=1+3i, if, for every ε > 0, there is a δ > 0 such that |(2 + i)z − (1 + 3i)| < ε whenever 0 < |z − (1 + i)| < δ. Proving that the limit exists requires that we find an appropriate value of δ for a given value of ε. In other words, for a given value of ε we must find a positive number δ with the property that if 0 < |z − (1 + i)| < δ, then |(2 + i)z − (1 + 3i)| < ε. One way of finding δ is to “work backwards.” The idea is to start with the inequality:

\left|(2+i)z-(1+3i)\right| \lt \varepsilon     (4)

and then use properties of complex numbers and the modulus to manipulate this inequality until it involves the expression |z − (1 + i)|. Thus, a natural first step is to factor (2 + i) out of the left-hand side of (4):

\left|2  +  i\right|. \left|z-\frac{1  +  3i}{2  +  i}\right| \lt \varepsilon      (5)

Because \left|2  +  i\right|=\sqrt{5} and  \frac{1  +  3i}{2  +  i} = 1 + i, (5) is equivalent to:

\sqrt{5} .\left|z  –  (1  +  i)\right|\lt \varepsilon   \text{or}    \left|z-(1  +  i)\right|\lt \frac{\varepsilon }{\sqrt{5} } .     (6)

Thus, (6) indicates that we should take \delta =\varepsilon /\sqrt{5} . Keep in mind that the choice of δ is not unique. Our choice of \delta =\varepsilon /\sqrt{5} is a result of the particular algebraic manipulations that we employed to obtain (6). Having found δ we now present the formal proof that \underset{z\rightarrow 1+i}{lim} (2+i)z=1+3i that does not indicate how the choice of δ was made:

Given \varepsilon \gt 0, let \delta =\varepsilon /\sqrt{5} . If 0\lt \left|z-(1+i)\right|\lt \delta, then we have \left|z-(1+i)\right|\lt \varepsilon /\sqrt{5}. Multiplying both sides of the last inequality by \left|2+i\right| =\sqrt{5} we obtain:

Therefore, \left|(2+i)z-(1+3i)\right| \lt\varepsilon whenever 0\lt \left|z-(1+i)\right|\lt\delta. So,
according to Definition 2.8, we have proven that \underset{z\rightarrow 1+i}{lim} (2+i)z=1+3i.