Ibrahim and Darus Journal of Inequalities and Applications 2013, 2013:51

Theorem 1. The function $S_k(z)$ satisfies $$\frac{1 - |a_k| r^{k-1}}{1 - \frac{|a_k|}{k} r^{k-1}} \le \Re\left(\frac{z S_k'(z)}{S_k(z)}\right) \le \frac{1 + |a_k| r^{k-1}}{1 + \frac{|a_k|}{k} r^{k-1}}$$ (3) for $$0 \le r < \sqrt[k-1]{\frac{k}{|a_k|}} \le 1, \quad |a_k| \ne 0.$$ Furthermore, $S_k(z) \in S^*(\alpha)$ for $$0 \le r < \sqrt[k-1]{\frac{1-\alpha}{(1-\alpha/k)|a_k|}} \le 1, \quad |a_k| \ne 0.$$ Proof Noting that $$\frac{zS_k'(z)}{S_k(z)} = k - \frac{(k-1)}{1 + \frac{a_k}{k}z^{k-1}},$$ it follows that for $\cos \theta \rightarrow 1$ , we obtain $$\Re\left(\frac{zS'_k(z)}{S_k(z)}\right) = k - (k-1)\frac{1 + \frac{|a_k|}{k}\cos\theta r^{k-1}}{1 + 2\frac{|a_k|}{k}r^{k-1}\cos\theta + (\frac{|a_k|}{k})^2 r^{2(k-1)}}$$ $$\leq \frac{1 + |a_k|r^{k-1}}{1 + \frac{|a_k|}{k}r^{k-1}}.$$ Moreover, we also observe that $$\Re\left(\frac{zS'_k(z)}{S_k(z)}\right) \ge \frac{1 - |a_k| r^{k-1}}{1 - \frac{|a_k|}{k} r^{k-1}}.$$ Now assume that $$\frac{1 - |a_k| r^{k-1}}{1 - \frac{|a_k|}{k} r^{k-1}} > \alpha$$ for $$0 \le r < \sqrt[k-1]{\frac{1-\alpha}{(1-\alpha/k)|a_k|}} \le 1, \quad |a_k| \ne 0.$$ This completes the proof. Remark 2 For example, the values $\alpha = 0.5$ , k = 2 and $|a_k| = 1$ imply the radius of starlikeness of $S_k(z)$ is r = 0.8164965..., and for the same values, the radius of starlikeness of the ordinary partial sums $f_k(z) = z + a_k z^k$ is r = 0.577350... (see [2]). Next, we derive the radius of convexity.

Theorem 3. The function $S_k(z)$ satisfies $$\frac{1 - k|a_k|r^{k-1}}{1 - |a_k|r^{k-1}} \le \Re\left(1 + \frac{zS_k''(z)}{S_k'(z)}\right) \le \frac{1 + k|a_k|r^{k-1}}{1 + |a_k|r^{k-1}} \tag{4}$$ for $$0 \le r < \sqrt[k-1]{\frac{1}{|a_k|}} \le 1, \quad |a_k| \ne 0.$$ Furthermore, $S_k(z) \in C(\alpha)$ for $$0 \le r < \sqrt[k-1]{\frac{1-\alpha}{(k-\alpha)|a_k|}} \le 1, \quad |a_k| \ne 0.$$ Proof A computation gives $$1 + \frac{zS_k''(z)}{S_k'(z)} = k - \frac{(k-1)}{1 + a_k z^{k-1}}.$$ Therefore, for $\cos \theta \rightarrow 1$ , we obtain $$\Re\left(1+\frac{zS_k''(z)}{S_k'(z)}\right) = k - (k-1)\frac{1+|a_k|\cos\theta r^{k-1}}{1+2|a_k|r^{k-1}\cos\theta + |a_k|^2r^{2(k-1)}}$$ $$\leq \frac{1+k|a_k|r^{k-1}}{1+|a_k|r^{k-1}}.$$ Moreover, we impose $$\Re\left(1 + \frac{zS_k''(z)}{S_k'(z)}\right) \ge \frac{1 - k|a_k|r^{k-1}}{1 - |a_k|r^{k-1}}.$$ Now, consider that $$\frac{1 - k|a_k|r^{k-1}}{1 - |a_k|r^{k-1}} > \alpha$$ for $$0 \le r < \sqrt[k-1]{\frac{1-\alpha}{(k-\alpha)|a_k|}} \le 1, \quad |a_k| \ne 0.$$ This completes the proof. Remark 4 In view of Theorem 3, for example, the values $\alpha = 0.5$ , k = 2 and $|a_k| = 1$ pose the radius of convexity of $S_k(z)$ is r = 0.577350... and for the same values, the radius of convexity of the ordinary partial sums $f_k(z) = z + a_k z^k$ is r = 0.4082... (see [2]). Next, we assume special ordinary partial sums depending so that their coefficients satisfy the relation $|a_n| \le (\frac{k-n+1}{k})$ .

Theorem 5. Assume the partial sum $$f_3(z) = z + \frac{k-1}{k}z^2 + \frac{k-2}{k}z^3$$ , $k \ge 2$ . Then the function $f_3(z) \in S^*(\frac{1}{2})$ . Proof We consider $\alpha$ such that $$\Re\left(\frac{zf_3'(z)}{f_3(z)}\right) = \Re\left(3 - \frac{2 + \frac{k-1}{k}z}{1 + \frac{k-1}{k}z + \frac{k-2}{k}z^2}\right) > \alpha.$$ This implies that $$\Re\left(\frac{2+\frac{k-1}{k}z}{1+\frac{k-1}{k}z+\frac{k-2}{k}z^2}\right) < 3-\alpha,$$ that is, $$\Re\left(\frac{1-\frac{k-1}{k}z^2}{1+\frac{k-1}{k}z+\frac{k-2}{k}z^2}\right) = \frac{1-\frac{k-1}{k}r^2(2\cos^2\theta-1)}{1+\frac{k-1}{k}r\cos\theta+\frac{k-2}{k}r^2(2\cos^2\theta-1)} < 2-\alpha.$$ By letting $t = \cos \theta$ , we define the function g(t) as follows: $$g(t) = \frac{1 - \frac{k-1}{k}r^2(2t^2 - 1)}{1 + \frac{k-1}{k}rt + \frac{k-2}{k}r^2(2t^2 - 1)}.$$ Logarithmic derivative of g(t) yields $$\begin{split} \frac{g'(t)}{g(t)} &= - \left\{ \frac{(4t\frac{k-1}{k}r^2)[1 + \frac{k-1}{k}rt + \frac{k-2}{k}r^2(2t^2 - 1)] + (\frac{k-1}{k}r + 4\frac{k-2}{k}r^2t)[1 - \frac{k-1}{k}r^2(2t^2 - 1)]}{[1 - \frac{k-1}{k}r^2(2t^2 - 1)][1 + \frac{k-1}{k}rt + \frac{k-2}{k}r^2(2t^2 - 1)]} \right\} \\ &:= - \left\{ \frac{h(t)}{[1 - \frac{k-1}{k}r^2(2t^2 - 1)][1 + \frac{k-1}{k}rt + \frac{k-2}{k}r^2(2t^2 - 1)]} \right\} \\ &= - \frac{At^2 + Bt + C}{[1 - \frac{k-1}{k}r^2(2t^2 - 1)][1 + \frac{k-1}{k}rt + \frac{k-2}{k}r^2(2t^2 - 1)]}, \end{split}$$ where $$A = 2r^{3} \left(\frac{k-1}{k}\right)^{2},$$ $$B = 4r^{2} \left[\frac{k-1}{k} + \frac{k-2}{k}\right],$$ $$C = \frac{k-1}{k} r \left[1 + \frac{k-1}{k} r^{2}\right].$$ Now, for all $k \ge 2$ and $r \to 1$ , the function h(t) has unique real negative zeros in the interval $(-\frac{1}{2},0)$ . This leads to the fact that g'(t) has unique positive real zeros for all $k \ge 2$ distributed in the interval $(0, \frac{1}{2})$ . Therefore, we will calculate $\alpha$ in $t \in [\frac{1}{2}, 1)$ . It is easy to check that g(t) is decreasing for $r \to 1$ in the interval $[\frac{1}{2}, 1)$ . Moreover, we have $$\lim_{k \to \infty} g\left(\frac{1}{2}\right) = \lim_{k \to \infty} \frac{1 + \frac{k-1}{2k}}{1 + \frac{k-1}{2k} - \frac{k-2}{2k}} = \frac{3}{2}.$$ We conclude that for $t \in [\frac{1}{2}, 1)$ , $$g(t) < g\left(\frac{1}{2}\right) \le \frac{3}{2} = 2 - \alpha,$$ thus $\alpha = \frac{1}{2}$ . This completes the proof. By letting k = 3 in Theorem 5, we have the following result. Corollary 6 The Cesáro partial sums $$\sigma_3(z) = z + \frac{2}{3}z^2 + \frac{1}{3}z^3, \quad z \in U,$$ of the function $f(z) = \frac{z}{1-z}$ are starlike of order $\alpha = \frac{1}{5}$ .

Theorem 7. Assume the partial sum $f_3(z)$ as in Theorem 5. Then the function $f_3(z) \in C(\frac{1}{5})$ . Proof We consider $\alpha$ such that $$\Re\left(1+\frac{zf_3''(z)}{f_3'(z)}\right)=\Re\left(3-\frac{2(\frac{k-1}{k}z+1)}{1+2\frac{k-1}{k}z+3\frac{k-2}{k}z^2}\right)>\alpha.$$ This implies that $$\Re\left(\frac{\frac{k-1}{k}z+1}{1+2\frac{k-1}{k}z+3\frac{k-2}{k}z^2}\right) < \frac{3-\alpha}{2},$$ therefore, a computation gives $$\Re\left(\frac{\frac{k-1}{k}z+1}{1+2\frac{k-1}{k}z+3\frac{k-2}{k}z^2}\right) = \frac{1}{2} + \Re\left(\frac{\frac{1}{2}(1-3\frac{k-2}{k}z^2)}{1+2(\frac{k-1}{k})z+3(\frac{k-2}{k})z^2}\right),$$ thus $$\frac{\frac{1}{2}(1-3\frac{k-2}{k}r^2(2\cos^2\theta-1))}{1+2(\frac{k-1}{k})r\cos\theta+3(\frac{k-2}{k})(2\cos^2\theta-1)}<\frac{2-\alpha}{2}.$$ By putting $t = \cos \theta$ , we define the function G(t) as follows: $$G(t) = \frac{\frac{1}{2}(1 - 3\frac{k-2}{k}r^2(2t^2 - 1))}{1 + 2(\frac{k-1}{k})rt + 3r^2(\frac{k-2}{k})(2t^2 - 1)}.$$ Logarithmic derivative of G(t) yields $$\begin{split} \frac{G'(t)}{G(t)} &= - \left\{ \frac{[12r^2\frac{k-2}{k}t][1+2(\frac{k-1}{k})rt+3r^2(\frac{k-2}{k})(2t^2-1)]}{[1-3\frac{k-2}{k}r^2(2t^2-1)][1+2(\frac{k-1}{k})rt+3r^2(\frac{k-2}{k})(2t^2-1)]} \right. \\ &+ \frac{[12r^2\frac{k-2}{k}t+2r\frac{k-1}{k}][1-3\frac{k-2}{k}r^2(2t^2-1)]}{[1-3\frac{k-2}{k}r^2(2t^2-1)][1+2(\frac{k-1}{k})rt+3r^2(\frac{k-2}{k})(2t^2-1)]} \right\} \\ &:= - \left\{ \frac{H(t)}{[1-3\frac{k-2}{k}r^2(2t^2-1)][1+2(\frac{k-1}{k})rt+3r^2(\frac{k-2}{k})(2t^2-1)]} \right\} \\ &= -\frac{At^2+Bt+C}{[1-3\frac{k-2}{k}r^2(2t^2-1)][1+2(\frac{k-1}{k})rt+3r^2(\frac{k-2}{k})(2t^2-1)]}, \end{split}$$ where $$A = 12r^{3} \frac{(k-1)(k-2)}{k^{2}},$$ $$B = 12r^{2} \frac{k-2}{k} \left[ 2 + 3r^{2} \left( \frac{k-2}{k} \right) \right],$$ $$C = 2r \frac{k-1}{k} \left[ 1 + 3r^{2} \left( \frac{k-2}{k} \right) \right].$$ Now, for all $k \ge 3$ and $r \to 1$ , the function H(t) has unique real negative zeros in the interval $[-\frac{1}{2},0)$ . This leads to the fact that G'(t) has unique positive real zeros for all $k \ge 3$ in the interval $(0,\frac{1}{2}]$ . So, we calculate $\alpha$ in the interval $t \in (\frac{1}{2},1)$ . A computation yields G(t) is decreasing for $r \to 1$ in the interval $t \in (\frac{1}{2},1)$ . Thus, we have $$\lim_{t \to \infty} G(t) < \frac{9}{10} = \frac{2 - \alpha}{2}, \quad 1 > t > 0.5,$$ which implies $\alpha = \frac{1}{5}$ . This completes the

Theorem 8. Assume the Cesáro partial sum $$\sigma_3(z) = z + \frac{4}{3}z^2 + z^3$$ of the function $$f(z) = \frac{z}{(1-z)^2} = z + 2z^2 + 3z^3 + \cdots$$ Then the function $\sigma_3(z) \in \mathcal{C}(\frac{1}{2})$ for all 0.2 < r < 0.5. Proof We consider $\alpha$ such that $$\Re\left(1+\frac{z\sigma_3''(z)}{\sigma_3'(z)}\right)=\Re\left(3-\frac{2(\frac{4}{3}z+1)}{1+\frac{8}{3}z+3z^2}\right)>\alpha.$$ This implies that $$\Re\left(\frac{\frac{4}{3}z+1}{1+\frac{8}{3}z+3z^2}\right)<\frac{3-\alpha}{2},$$ therefore, a computation gives $$\Re\left(\frac{\frac{4}{3}z+1}{1+\frac{8}{3}z+3z^2}\right)=\frac{1}{2}+\Re\left(\frac{\frac{1}{2}(1-3z^2)}{1+\frac{8}{3}z+3z^2}\right),$$ thus $$\frac{1 - 3r^2(2\cos^2\theta - 1)}{1 + \frac{8}{2}r\cos\theta + 3(2\cos^2\theta - 1)} < 2 - \alpha.$$ By putting $t = \cos \theta$ , we define the function j(t) as follows: $$J(t) = \frac{1 - 3r^2(2t^2 - 1)}{1 + \frac{8}{3}rt + 3r^2(2t^2 - 1)}.$$ Logarithmic derivative of j(t) yields $$\begin{split} \frac{J'(t)}{J(t)} &:= -\left\{\frac{\hbar(t)}{[1-3r^2(2t^2-1)][1+\frac{8}{3}rt+3r^2(2t^2-1)]}\right\} \\ &= -\frac{16r^3t^2+24r^2t+\frac{8}{3}r(1+3r^2)}{[1-3r^2(2t^2-1)][1+\frac{8}{3}rt+3r^2(2t^2-1)]}. \end{split}$$ The function $\hbar(t)$ has a unique real negative zero in the interval $t \in (-1,0)$ for all 0.2 < r < 0.5 which is around $t \simeq -\frac{1}{2}$ . This leads to the fact that J'(t) has a unique positive real zero in the interval (0,1) around $t \simeq \frac{1}{2}$ . A computation yields J(t) is decreasing in the interval $t \in (\frac{1}{2},1)$ and assuming its maximums at t = 0.5 and t = 0.5. Thus, we have $$\lim_{t \to 0.5, t \to 0.5} J(t) < \frac{3}{2} = 2 - \alpha,$$ <span id="page-7-0"></span>which implies $\alpha = \frac{1}{2}$ . This completes the proof. <span id="page-7-1"></span>Note that some other results related to partial sums can be found in [11–15].