In our paper, which is “Complex symmetric weighted composition operators on H2(D)”, we made the incorrect statement in page 331, lines 13 and 14 as follows: From Eq. (6) and Lemma 4.8, we obtain a complex symmetric composition operator which is not hyponormal. On the other hand, we realized that a0λ2−(1+a02−a1)λ+a0=0 since φ(λ)=λ. Thus, since λ is real, Eq. (6) in  becomes φ˜(z)=−(|λ|2−a0(λ+λ‾)+a02−a1)z(a0λ‾2−(1+a02−a1)λ‾+a0)z+(a02−a1)|λ|2−a0(λ+λ‾)+1=−(|λ|2−a0(λ+λ‾)+a02−a1)(a02−a1)|λ|2−a0(λ+λ‾)+1z=βz,z∈D where β=−(|λ|2−a0(λ+λ‾)+a02−a1)(a02−a1)|λ|2−a0(λ+λ‾)+1 and |β|≤1, since φ˜ is the composition of analytic maps of the unit disk into itself. Therefore, Cφ˜ is normal. Thus we replace the above incorrect statement (in ) by the following: From Eq. (6) and Lemma 4.8, we obtain a complex symmetric composition operator which is normal. Waleed Noor also commented on the above incorrect statement as follows: The weighted composition operator Wψλ,bλ (notation as in [2, Corollary 3.5]) is unitary and Hermitian [1, Theorem 3.1]. Since λ is real, it is easy to see that Wψλ,bλ commutes with (Jf)(z)=f(z‾)‾, i.e., Wψλ,bλ is J-symmetric. It follows that Wψλ,bλ⁎JWψλ,bλ=J and Cφ˜ is J-symmetric if and only if it is normal. In addition, Waleed Noor pointed out some mistakes in the calculation of Example 3.6 in page 331 of . Now we give the correct calculation in the following example.Example 3.6 If φ(z)=−122+z and ψ(z)=222+z (i.e., a0=−122, a1=18, and b=12), then we know that φ is an analytic selfmap of D from Lemma 4.8. Thus, Wψ,φ is complex symmetric with conjugation J by Theorem 3.3. Moreover, φ has the fixed point λ=−2+1∈D with Im(λ)=0. Under the same notations as in Corollary 3.5 in , we get from Eq. (6) that Cφ˜ is complex symmetric with conjugation (Wψλ,bλ)⁎J(Wψλ,bλ), where φ˜(z)=(3−22)z. So, it is obvious that Cφ˜ is normal. Hence we complete the correction for our paper .