1 Introduction

Denote by \({\mathcal {A}}\) a class of functions

$$\begin{aligned} g(z)=z+\sum _{n=2}^{\infty }a_nz^n \end{aligned}$$
(1.1)

which are analytic in the unit disk \({\mathbb {D}}=\{z\in {\mathbb {C}}: |z|<1\}\). Let \({\mathcal {S}}\) be a subclass of \({\mathcal {A}}\) consisting of all univalent functions in \({\mathbb {D}}\). Furthermore, let C(r, g) be the closed curve which is the image of \(|z|=r<1\) under the mapping \(w=g(z)\). Denote by L(r, g) the length of C(r, g) and let A(r, g) be the area enclosed by C(r, g).

If \(g(z)\in {\mathcal {A}}\) satisfies

$$\begin{aligned} {{\mathfrak {R}}{\mathfrak {e}}} \left\{ \frac{zg'(z)}{g(z)}\right\} >0,\quad z\in {\mathbb {D}}, \end{aligned}$$

then g(z) is said to be starlike with respect to the origin in \({\mathbb {D}}\) and we write \(g(z)\in {\mathcal {S}}^*\). It is known that \({\mathcal {S}}^*\subset {\mathcal {S}}\). We say that \(f(z)\in {{\mathcal {A}}}\) is convex of order \(\alpha \), where \(0\le \alpha <1\), provided

$$\begin{aligned} {{\mathfrak {R}}{\mathfrak {e}}}\left\{ 1+\frac{zf''(z)}{f'(z)}\right\} >\alpha ,\quad z\in {\mathbb {D}}. \end{aligned}$$

The class of convex functions of order \(\alpha \) has been introduced by Robertson in [14]. In what follows this class is denoted \({\mathcal {K}}(\alpha )\). It is well known that \(\mathcal K(\alpha )\subset {\mathcal {S}}\) for all \(\alpha \), \(0\le \alpha <1\). A function \(f(z)\in {\mathcal {A}}\) is said to be close-to-convex if it satisfies the following condition

$$\begin{aligned} {{\mathfrak {R}}{\mathfrak {e}}} \left\{ \frac{zf'(z)}{e^{i\alpha }g(z)}\right\} >0,\ z\in {\mathbb {D}} \end{aligned}$$

for some \(g(z)\in {\mathcal {S}}^*\) and \(\alpha \in (-\pi /2,\pi /2)\). The class of close-to-convex functions is denoted by \({\mathcal {C}}\). Let us recall that \({\mathcal {C}}\subset {\mathcal {S}}\). An univalent function \(f(z)\in {\mathcal {S}}\) belongs to \({\mathcal {C}}\) if and only if a complement E of the image-region \(F=\left\{ f(z): |z|<1\right\} \) is a union of rays that are disjoint (except that the origin of one ray may belong to another ray).

It is known that, for any \(r\in (0,1)\), the inequality

$$\begin{aligned} L(r,f)\le L\left( r,\frac{z}{(1-z)^2}\right) \end{aligned}$$

holds for every \(f(z)\in {\mathcal {S}}^*\) (see [6, p.65]) and every \(f(z)\in {\mathcal {C}}\) (see [1]). Moreover, Marx proved in [6, p.65] that

$$\begin{aligned} L(r,f)\le L\left( r,\frac{z}{1-z}\right) =\frac{2\pi r}{1-r^2} \end{aligned}$$

for all \(r\in (0,1)\) and every \(f(z)\in {\mathcal {K}}(0)\). In [2], Eenigenburg extended this result to the class \({\mathcal {K}}(\alpha )\), where \(\alpha \in [0,1)\). Namely, for all \(r\in (0,1)\) and every \(f(z)\in {\mathcal {K}}(\alpha )\), we have

$$\begin{aligned} L(r,f)&\le L\left( r,-\log (1-z)\right) ,\quad \alpha =1/2,\\ L(r,f)&\le L\left( r,[1-(1-z)^{2\alpha -1}/(2\alpha -1)\right) ,\quad \alpha \ne 1/2. \end{aligned}$$

In the other sets, the extremal functions are more complicated or the length problem is unsolved as is in the class \({\mathcal {S}}\).

Let

$$\begin{aligned} M(r,g)=\max _{|z|=r<1}|g(z)|. \end{aligned}$$

In [5] F. R. Keogh has proved the following result.

Theorem 1.1

[5] Assume that \(g(z)\in {\mathcal {S}}^*\). Then we have

$$\begin{aligned} L(r,g)={\mathcal {O}}\left( M(r,g)\log \frac{1}{1-r}\right) \quad \textrm{as}\quad r\rightarrow 1, \end{aligned}$$
(1.2)

where \({\mathcal {O}}\) denotes the Landau’s symbol.

Let us note that in [15] D. K. Thomas extended this result to the class of bounded close-to-convex functions. In [16] D. K. Thomas has shown also that

Theorem 1.2

[16, Th.1] If \(g(z)\in {\mathcal {S}}^*\), then

$$\begin{aligned} 2\sqrt{\pi A(r,g)}\le L(r,g)\le 2\sqrt{\pi A(r,g)}\left( 1+\log \frac{1+r}{1-r}\right) \quad \textrm{as}\quad r\rightarrow 1. \end{aligned}$$

In particular \(L(r,g)\sim 2\sqrt{\pi A(r,g)}\) as \(r\rightarrow 0\). In [16] D. K. Thomas has shown that

Theorem 1.3

[16] If \(g(z)\in {\mathcal {C}}\), then (1.2) holds, and if

$$\begin{aligned} M(r,g)\le (1-r)^{-\alpha } \end{aligned}$$

for some \(\alpha \in (0,2]\), then

$$\begin{aligned} L(r,g)=k(\alpha )(1-r)^{-\alpha }. \end{aligned}$$

where \(k(\alpha )\) depends only on \(\alpha \).

In [13] Ch. Pommerenke extended the first part of Theorem 1.3 as follows

Theorem 1.4

[13] If \(g(z)\in {\mathcal {C}}\), then

$$\begin{aligned} L(r,g)=\mathcal O\left\{ M(r,g)\left( \log \frac{1}{1-r}\right) ^{5/2}\right\} \quad \textrm{as}\quad r\rightarrow 1. \end{aligned}$$

In [4] W. F. Hayman has shown that, if \(g(z)\in \mathcal S^*\) and \(A(r,g)<A\) for some constant A, then

$$\begin{aligned} L(r,g)={\mathcal {O}} \left( \log \frac{1}{1-r}\right) \quad \textrm{as}\quad r\rightarrow 1. \end{aligned}$$
(1.3)

Hayman gave an example a bounded starlike function g(z) satisfying

$$\begin{aligned} \limsup _{r\rightarrow 1}\frac{L(r,g)}{\log \frac{1}{1-r}}>0. \end{aligned}$$

This example clearly shows that the \({\mathcal {O}}\) in (1.3) cannot in general be replaced by small \( o\).

The similar problems have been considered in [7, 8]. The following result has been proved in [8].

Theorem 1.5

[8] If \(zg'(z)\in {\mathcal {S}}^*\), then

$$\begin{aligned} L(r,g)={\mathcal {O}}\left\{ A(r,g)\log \frac{1}{1-r}\right\} ^{1/2}\quad \textrm{as}\quad r\rightarrow 1. \end{aligned}$$

Note that the above hypothesis \(zg'(z)\in {\mathcal {S}}^*\) is equivalent to that g is convex univalent in \({\mathbb {D}}\). Some related length problems were considered in [9,10,11]. In [15], D. K. Thomas considered L(r, g) for the class of bounded close-to-convex functions and asked the following question. Does there exist a starlike function g(z) for which

$$\begin{aligned} \liminf _{r\rightarrow 1}\frac{L(r,g)}{M(r,g)\log \frac{1}{1-r}}>0 \end{aligned}$$

or

$$\begin{aligned} \liminf _{r\rightarrow 1}\frac{L(r,g)}{\sqrt{A(r,g)}\log \frac{1}{1-r}}>0? \end{aligned}$$
(1.4)

In [12] a negative answer to the open problem (1.4) was given under some additional condition. Some related problems were considered in [9, 10].

Theorem 1.6

[9, Th.1.8] If \(g(z)\in {\mathcal {S}}\), then

$$\begin{aligned} M(r,g)\le 4\sqrt{\frac{A(r,g)}{\pi }\log \frac{3}{1-r}}\quad (|z|=r<1). \end{aligned}$$

Therefore, we have

$$\begin{aligned} M(r,g)={\mathcal {O}}\left\{ A(r,g)\log \frac{1}{1-r}\right\} ^{1/2}\quad \textrm{as}\quad r\rightarrow 1. \end{aligned}$$
(1.5)

In [17] M. Tsuji has shown that

Theorem 1.7

[17, p. 227] If \(f(z)=u(z)+iv(z)\) is analytic in \(|z|\le R\), then

$$\begin{aligned} f(z)=\frac{1}{2\pi }\int _0^{2\pi }u(Re^{i\phi })\frac{Re^{i\phi }+z}{Re^{i\phi }-z}\textrm{d}\phi +iv(0) \end{aligned}$$

for all z, \(|z|<R\).

Moreover, if \(|z|<R\), \(v(0)=0\), then

$$\begin{aligned} |f(z)|\le \frac{1}{2\pi }\int _0^{2\pi }|u(Re^{i\varphi })|\left| \frac{Re^{i\varphi }+z}{Re^{i\varphi }-z}\right| \textrm{d}\varphi . \end{aligned}$$
(1.6)

It is known that

$$\begin{aligned} {\mathfrak {R}}{\mathfrak {e}}\left\{ \frac{Re^{i\phi }+z}{Re^{i\phi }-z}\right\} =\frac{R^2-r^2}{R^2+r^2-2Rr\cos (\phi -\theta )}, \end{aligned}$$

for all \(r<R\), where \(z=re^{i\theta }\), and

$$\begin{aligned} \int _0^{2\pi } \frac{R^2-r^2}{R^2+r^2-2Rr\cos (\phi -\theta )}\textrm{d}\theta =2\pi . \end{aligned}$$
(1.7)

Moreover, from [17, p.226], we have

$$\begin{aligned} \int _{-\pi }^{\pi } \frac{R^2-r^2}{R^2+r^2-2Rr\cos (\phi -\theta )}\textrm{d}\theta =2\pi \end{aligned}$$
(1.8)

and

$$\begin{aligned} \int _{\alpha }^{\alpha +2\pi } \frac{R^2-r^2}{R^2+r^2-2Rr\cos (\phi -\theta )}\textrm{d}\theta =2\pi \end{aligned}$$
(1.9)

for all real \(\alpha \). It is a purpose of this paper to prove, using a modified method, that a result related to (1.2), holds also under some another assumptions on g(z).

2 Main results

Theorem 2.1

If f(z) and \(zf'(z)/f(z)\) are analytic in \(|z|\le R\), with \({\mathfrak {I}}{\mathfrak {m}}\{zf'(z)/f(z)\}|_{z=0}=0\), then

$$\begin{aligned} L(r,f)\le M(r,f)V(R)\left( \frac{2}{\pi }\log \frac{R+r}{R-r}+1\right) , \end{aligned}$$
(2.1)

for all r, \(0<r<R\), where

$$\begin{aligned} V(R)=\int _0^{2\pi }\left| \frac{\textrm{d}\arg \{f(Re^{i\phi })\}}{\textrm{d}\phi }\right| \textrm{d}\phi . \end{aligned}$$
(2.2)

Proof

Let \(z=r e^{i\theta }\) and let

$$\begin{aligned} B(R,\phi )=\mathfrak {Re}\left\{ \frac{Re^{i\phi }f'(Re^{i\phi })}{f(Re^{i\phi })}\right\} . \end{aligned}$$

Applying (1.6) for \(zf'(z)/f(z)\) and the hypothesis of Theorem 2.1, we have

$$\begin{aligned} L(r,f)&=\int _0^{2\pi }|zf'(z)|\textrm{d}\theta \nonumber \\&=\int _0^{2\pi }\left| \frac{zf'(z)}{f(z)}\right| |f(z)|\textrm{d}\theta \nonumber \\&\le M(r,f)\int _0^{2\pi }\left| \frac{zf'(z)}{f(z)}\right| \textrm{d}\theta \nonumber \\&= \frac{M(r,f)}{2\pi }\int _0^{2\pi }\left| \int _0^{2\pi }{\mathfrak {R}}{\mathfrak {e}}\left\{ \frac{Re^{i\phi }f'(Re^{i\phi })}{f(Re^{i\phi })}\right\} \frac{R e^{i\theta }+z}{Re^{i\theta }-z}\textrm{d}\phi \right| \textrm{d}\theta \nonumber \\&\le \frac{M(r,f)}{2\pi }\int _0^{2\pi }\int _0^{2\pi }\left| B(R,\phi )\right| \left| \frac{R e^{i\phi }+r e^{i\theta }}{Re^{i\phi }-r e^{i\theta }}\right| \textrm{d}\phi \textrm{d}\theta \nonumber \\&= \frac{M(r,f)}{2\pi }\int _0^{2\pi }\int _0^{2\pi }\left| B(R,\phi )\right| \left| \frac{R^2-r^2-i2Rr\sin (\phi -\theta )}{R^2+r^2-2Rr\cos (\phi -\theta )}\right| \textrm{d}\phi \textrm{d}\theta \nonumber \\&\le \frac{M(r,f)}{2\pi }\int _0^{2\pi }\int _0^{2\pi }\left| B(R,\phi )\right| \frac{R^2-r^2}{R^2+r^2-2Rr\cos (\phi -\theta )}\textrm{d}\theta \textrm{d}\phi \nonumber \\&~ + \frac{M(r,f)}{2\pi }\int _0^{2\pi }\int _0^{2\pi }\left| B(R,\phi )\right| \left| \frac{-2Rr\sin (\phi -\theta )}{R^2+r^2-2Rr\cos (\phi -\theta )}\right| \textrm{d}\theta \textrm{d}\phi \nonumber \\&= M(r,f)\int _0^{2\pi }\int _0^{2\pi }\left| B(R,\phi )\right| \textrm{d}\phi \nonumber \\&~ + \frac{M(r,f)}{2\pi }\int _0^{2\pi }\int _0^{2\pi }\left| B(R,\phi )\right| \left| \frac{-2Rr\sin (\phi -\theta )}{R^2+r^2-2Rr\cos (\phi -\theta )}\right| \textrm{d}\theta \textrm{d}\phi \end{aligned}$$
(2.3)

because of the equality (1.7). For the second integral (2.3), we have

$$\begin{aligned}&~\frac{M(r,f)}{\pi }\int _0^{2\pi }\int _{\phi }^{\phi +\pi }\left| B(R,\phi )\right| \frac{|2Rr\sin (\phi -\theta )|}{R^2+r^2-2Rr\cos (\phi -\theta )}\textrm{d}\theta \textrm{d}\phi \\&=\frac{M(r,f)}{\pi }\int _0^{2\pi }\left| B(R,\phi )\right| \left. \log (R^2+r^2-2Rr\cos (\phi -\theta ))\right| _{\phi }^{\phi +\pi }\textrm{d}\phi \\&=\frac{2M(r,f)}{\pi }\int _0^{2\pi }\left| B(R,\phi )\right| \log \frac{R+r}{R-r}\textrm{d}\phi \\&=\frac{2M(r,f)}{\pi }\log \frac{R+r}{R-r}\int _0^{2\pi }\left| B(R,\phi )\right| \textrm{d}\phi \\&=\frac{2M(r,f)}{\pi }\log \frac{R+r}{R-r}\int _0^{2\pi }\left| \frac{\textrm{d}\arg \{f(Re^{i\phi })\}}{\textrm{d}\phi }\right| \textrm{d}\phi . \end{aligned}$$

Therefore, from (2.3), we have

$$\begin{aligned} L(r,f)&\le M(r,f)V(R)\left( \frac{2}{\pi }\log \frac{R+r}{R-r}+1\right) , \end{aligned}$$

where V(R) has the form (2.2). \(\square \)

Remark

Putting \(R=\sqrt{r}<1\), we have

$$\begin{aligned} \frac{R+r}{R-r}=\frac{\sqrt{r}+r}{\sqrt{r}-r} =\frac{1+2\sqrt{r}+r}{1-r}\le \frac{1+3\sqrt{r}}{1-r}<\frac{4}{1-r}. \end{aligned}$$

Moreover, the function

$$\begin{aligned} G(R)=\frac{R+r}{R-r},\quad 0<R\le 1 \end{aligned}$$

is increasing, so

$$\begin{aligned} \frac{R+r}{R-r}<\frac{1+r}{1-r}. \end{aligned}$$

Therefore, if f(z) is analytic in \(|z|\le 1\), then (2.1) becomes

$$\begin{aligned} L(r,f)\le M(r,f)V(1)\left( \frac{2}{\pi }\log \frac{1+r}{1-r}+1\right) =M(r,f)V(1)\left( \frac{2}{\pi }\log \frac{2}{1-r}+1\right) \end{aligned}$$

for all r, \(0<r<1\).

Theorem 2.2

If f(z) is analytic in \(|z|\le R\), \(zf'(z)=u(z)+iv(z)\) and \(|u(z)|\le 1\) in \(|z|\le R\), \(v(0)=0\), then

$$\begin{aligned} L(f,r)\le 2\pi +4\log \frac{R+r}{R-r} \end{aligned}$$

for all r, \(0<r<R\).

Proof

Write \(z=r e^{i\theta }\). Then, we have

$$\begin{aligned} \frac{R e^{i\phi }+z}{R e^{i\phi }-z}&=\frac{R e^{i\phi }+r e^{i\theta }}{R e^{i\phi }-r e^{i\theta }}=\frac{R^2-r^2-i2Rr\sin (\phi -\theta )}{R^2+r^2-2Rr\cos (\phi -\theta )}. \end{aligned}$$

Applying (1.6) for \(zf'(z)=u(z)+iv(z)\), we get

$$\begin{aligned} |zf'(z)|&=\left| \frac{1}{2\pi }\int _0^{2\pi } u(Re^{i\phi }) \frac{Re^{i\phi }+z}{Re^{i\phi }-z}\textrm{d}\phi +iv(0)\right| \nonumber \\&\le \frac{1}{2\pi }\int _0^{2\pi } \left| u(Re^{i\phi })\right| \left| \frac{Re^{i\phi }+z}{Re^{i\phi }-z}\right| \textrm{d}\phi \nonumber \\&\le \frac{1}{2\pi }\int _0^{2\pi } \left| \frac{Re^{i\phi }+r e^{i\theta }}{Re^{i\phi }-r e^{i\theta }}\right| \textrm{d}\phi \nonumber \\&=\frac{1}{2\pi }\int _0^{2\pi }\left| \frac{R^2-r^2-i2Rr\sin (\phi -\theta )}{R^2+r^2-2Rr\cos (\phi -\theta )}\right| \textrm{d}\phi \nonumber . \end{aligned}$$

Thus, in view of (1.7)–(1.9), we obtain

$$\begin{aligned} L(r,f)&=\int _0^{2\pi }|zf'(z)|\textrm{d}\theta \nonumber \\&=\frac{1}{2\pi }\int _0^{2\pi } \left[ \int _0^{2\pi }\left| \frac{R^2-r^2-i2Rr\sin (\phi -\theta )}{R^2+r^2-2Rr\cos (\phi -\theta )}\right| \textrm{d}\theta \right] \textrm{d}\phi \nonumber \\&\le \frac{1}{2\pi }\int _0^{2\pi }\int _\phi ^{2\pi +\phi }\frac{R^2-r^2}{r^2+r^2-2Rr\cos (\phi -\theta )}\textrm{d}\theta \textrm{d}\phi \nonumber \\&~+ \frac{1}{2\pi }\int _0^{2\pi }\int _\phi ^{2\pi +\phi }\frac{2Rr|\sin (\phi -\theta )|}{R^2+r^2-2Rr\cos (\phi -\theta )}\textrm{d}\theta \textrm{d}\phi \nonumber \\&=\frac{1}{2\pi }\int _0^{2\pi }\left( 2\pi +2\int _\phi ^{\pi +\phi }\frac{2Rr\sin (\phi -\theta )}{R^2+r^2-2Rr\cos (\phi -\theta )}\textrm{d}\theta \right) \textrm{d}\phi \nonumber \\&=\frac{1}{2\pi }\int _0^{2\pi }\left( 2\pi +4\log \frac{R+r}{R-r}\right) \textrm{d}\phi \nonumber \\&=2\pi +4\log \frac{R+r}{R-r}. \end{aligned}$$

For \(R=1\) Theorem 2.2 becomes the following corollary.\(\square \)

Corollary 2.3

If f(z) is analytic in \(|z|\le 1\), \(zf'(z)=u(z)+iv(z)\) and \(|u(z)|\le 1\) in \(|z|\le 1\), \(v(0)=0\), then

$$\begin{aligned} L(f,r)\le 2\pi +4\log \frac{1+r}{1-r} \end{aligned}$$

for all r, \(0<r<1\).

Theorem 2.4

Let g(z) be of the form (1.1) and suppose that

$$\begin{aligned} \left| \frac{zg'(z)}{g(z)}\right| \le \left| \frac{1+z}{1-z}\right| ,\quad z\in {\mathbb {D}} \end{aligned}$$
(2.4)

and

$$\begin{aligned} |g(z)|\le \left| \frac{1+z}{1-z}\right| ^{\beta },\quad z\in {\mathbb {D}} \end{aligned}$$
(2.5)

for some \(\beta \), \(1<\beta \). Then we have

$$\begin{aligned} L(r,g)\le \frac{8\pi (1+r)^{\beta -2}}{(1-r)^\beta }. \end{aligned}$$

Proof

From (2.4) and (2.5), it follows that

$$\begin{aligned} L(r,g)&=\int _0^{2\pi }|zg'(z)|\textrm{d}\nu \\&=\int _0^{2\pi }\left| r e^{i\nu }g'(r e^{i\nu })\right| \textrm{d}\nu \\&=\int _0^{2\pi }\left| \frac{zg'(z)}{g(z)}\right| |g(z)|\textrm{d}\nu \\&\le \int _0^{2\pi }\left| \frac{1+z}{1-z}\right| \left| \frac{1+z}{1-z}\right| ^\beta \textrm{d}\nu \\&= \int _0^{2\pi }|1+z|^2\frac{1}{|1-z|^{2}}\left| \frac{1+z}{1-z}\right| ^{\beta -1}\textrm{d}\nu \\&\le \int _0^{2\pi }\frac{4}{|1-z|^{2}}\left( \frac{1+r}{1-r}\right) ^{\beta -1}\textrm{d}\nu \\&\le \frac{4(1+r)^{\beta -1}}{(1-r)^{\beta -1}}\int _0^{2\pi }\frac{1}{|1-z|^{2}}\textrm{d}\nu \\&= \frac{4(1+r)^{\beta -1}}{(1-r)^{\beta -1}}\int _0^{2\pi }\frac{1}{1-2r\cos \nu +r^2}\textrm{d}\nu . \end{aligned}$$

From (1.7), we have

$$\begin{aligned} \int _0^{2\pi }\frac{1}{1-2r\cos \nu +r^2}\textrm{d}\nu =\frac{2\pi }{1-r^2}, \end{aligned}$$

therefore, we obtain

$$\begin{aligned} L(r,g)&\le \frac{4(1+r)^{\beta -1}}{(1-r)^{\beta -1}} \frac{2\pi }{1-r^2}. \end{aligned}$$

This completes the proof of Theorem 2.4. \(\square \)

Lemma 2.5

[3, p. 280], [2, p. 491]

$$\begin{aligned} \int _{0}^{2\pi }\frac{\textrm{d}\nu }{|1-re^{i\nu }|^{\beta }} =\left\{ \begin{array}{ll} {\mathcal {O}}\left( (1-r)^{1-\beta }\right) &{}for\ the\ case\ 1<\beta ,\\ {\mathcal {O}}\left( \log \frac{1}{1-r}\right) &{}for\ the\ case\ \beta =1,\\ {\mathcal {O}}\left( 1\right) &{}for\ the\ case\ 0\le \beta <1, \end{array} \right. \end{aligned}$$

where \(0<r<1\), \(0\le \nu \le 2\pi \), \(0\le \beta \) and where \(\mathcal O\) is the Landau’s symbol.

Corollary 2.6

Let g(z) be of the form (1.1) and suppose that

$$\begin{aligned} \left| \frac{zg'(z)}{g(z)}\right| \le \frac{A}{|1-z|^{\alpha }}\quad z\in \mathbb D \end{aligned}$$
(2.6)

and

$$\begin{aligned} \left| g(z)\right| \le \frac{B}{|1-z|^{\beta }}\quad z\in \mathbb D \end{aligned}$$
(2.7)

for some real \(\alpha , \beta \), \(\alpha +\beta >0\). Then we have

$$\begin{aligned} L(r,g)= \left\{ \begin{array}{ll} AB~ {\mathcal {O}}\left( (1-r)^{1-\beta }\right) &{}for\ the\ case\ 1<\alpha +\beta ,\\ AB~ {\mathcal {O}}\left( \log \frac{1}{1-r}\right) &{}for\ the\ case\ \alpha +\beta =1,\\ AB~ {\mathcal {O}}\left( 1\right) &{}for\ the\ case\ 0\le \alpha +\beta <1, \end{array} \right. , \end{aligned}$$

as \(r\rightarrow 1\), \(0<r<1\).

Proof

From (2.6) and (2.7), we get

$$\begin{aligned} \left| zg'(z)\right| \le \frac{AB}{|1-z|^{\alpha +\beta }}\quad z\in \mathbb D. \end{aligned}$$

Then, we have

$$\begin{aligned} L(r,g)&=\int _0^{2\pi }\left| r e^{i\nu }g'(r e^{i\nu })\right| \textrm{d}\nu \\&\le \int _0^{2\pi }\frac{AB}{|1-r e^{i\nu }|^{\alpha +\beta }}\textrm{d}\nu . \end{aligned}$$

It suffices to apply Hayman’s lemma 2.5. \(\square \)

The next result is an extension of Theorem 1.5.

Theorem 2.7

If g(z) is of the form (1.1), then we have

$$\begin{aligned} M(r)= {\mathcal {O}}\left( A(r)\log \frac{1}{1-r}\right) , \end{aligned}$$
(2.8)

as \(r\rightarrow 1\), \(0<r<1\).

Proof

Note that

$$\begin{aligned} M(r)= & {} \max _{|z|=r<1}\left| \int _0^z g'(s)\textrm{d}s\right| \\= & {} \max _{|z|=r<1}\left| \int _0^r g'(\rho e^{i\nu })e^{i\nu }\textrm{d}\rho \right| \\= & {} \max _{|z|=r<1}\left| \int _0^r g'(\rho e^{i\nu })\textrm{d}\rho \right| . \end{aligned}$$

Hence, in view of (1.6), we get

$$\begin{aligned} M(r)= & {} \max _{|z|=r<1}\left| \frac{1}{2\pi }\int _0^r\int _0^{2\pi } {\mathfrak {R}}{\mathfrak {e}}g'(t e^{i\nu })\frac{t e^{i\phi } +\rho e^{i\nu }}{te^{i\phi }-\rho e^{i\nu }}\textrm{d}\phi \textrm{d}\rho \right| \\\le & {} \max _{|z|=r<1}\frac{1}{2\pi }\int _0^r\int _0^{2\pi } \left| g'(t e^{i\nu })\right| \left| \frac{te^{i\phi } +\rho e^{i\nu }}{te^{i\phi }-\rho e^{i\nu }}\right| \textrm{d}\phi \textrm{d}\rho , \end{aligned}$$

where \(0\le \rho \le r<t<1\). Then, applying Schwarz’s lemma, we obtain

$$\begin{aligned} M(r)\le & {} \max _{|z|=r<1}\left( \frac{1}{2\pi }\int _0^r\int _0^{2\pi } \left| g'(t e^{i\nu })\right| ^2\textrm{d}\phi \textrm{d}\rho \right) ^{1/2} \left( \int _0^r\int _0^{2\pi }\left| \frac{te^{i\phi } +\rho e^{i\nu }}{te^{i\phi }-\rho e^{i\nu }}\right| ^2\textrm{d}\phi \textrm{d}\rho \right) ^{1/2}\\\le & {} \max _{|z|=r<1}(I_1)^{1/2}(I_2)^{1/2}, \ \ \textrm{say}. \end{aligned}$$

Putting \(0<r_1<r\), and \(t=\sqrt{(1+\rho ^2)/2}\), we get

$$\begin{aligned} \rho \textrm{d}\rho =2\sqrt{\frac{1+\rho ^2}{2}}\textrm{d}t<2\textrm{d}t. \end{aligned}$$

Thus, we have

$$\begin{aligned} I_1= & {} \frac{1}{2\pi }\int _0^{r_1}\int _0^{2\pi } \left| g'(t e^{i\phi })\right| ^2\textrm{d}\phi \textrm{d}\rho +\frac{1}{2\pi r^2_1}\int _{\sqrt{(1+r_1^2)/2}}^{\sqrt{(1+r^2)/2}}\int _0^{2\pi } t\left| g'(t e^{i\phi })\right| ^2\textrm{d}\phi \textrm{d}t \\\le & {} C+\frac{1}{2\pi r^2_1}A\left( \sqrt{\frac{1+r^2}{2}}\right) \\= & {} C+\frac{1}{2\pi r^2_1}A\left( \sqrt{\frac{1+r^2}{2r^2}}r\right) \\= & {} {\mathcal {O}}(A(r))\ \ \ \textrm{as}\ \ r\rightarrow 1, \end{aligned}$$

where C is a bounded positive constant. On the other hand, letting \(t\rightarrow 1^{-}\), we have

$$\begin{aligned} I_2= & {} \int _0^r\int _0^{2\pi }\left| \frac{te^{i\phi } +\rho e^{i\nu }}{te^{i\phi }-\rho e^{i\nu }}\right| ^2\textrm{d}\phi \textrm{d}\rho \\\le & {} \int _0^{r}\int _0^{2\pi }\frac{4}{\left| te^{i\phi }-\rho e^{i\nu }\right| ^2}\textrm{d}\phi \textrm{d}\rho \\= & {} \int _0^{r}\int _0^{2\pi }\frac{4}{t^2-2\rho t\cos (\phi -\nu )+\rho ^2}\textrm{d}\phi \textrm{d}\rho . \end{aligned}$$

Using (1.7), we get

$$\begin{aligned} I_2\le & {} 8\pi \int _0^{r}\frac{1}{t^2-\rho ^2}\textrm{d}\rho \\= & {} \frac{4\pi }{t}\int _0^{r}\left( \frac{1}{t+\rho }+\frac{1}{t-\rho }\right) \textrm{d}\rho \\= & {} \frac{4\pi }{t}\log \frac{t+r}{t-r}\rightarrow \mathcal O\left( \log \frac{1}{1-r} \right) \ \ \ \textrm{as}\ \ r\rightarrow 1. \end{aligned}$$

This completes the proof of (2.8). \(\square \)

Remark

In the above theorem we do not suppose that g(s) is univalent in \(|z|<1\) and therefore, L(r) and A(r) are not necessarily the length of the image curve C(r) and the area enclosed by the image curve C(r) which is the image curve of the circle under the mapping \(w=g(z)\).