A Zverev, Nishikori, Coric, Fognini, Medvedev enter 2nd round of AO 2019

Speed Post News Network

Patna : World No 4 and winner of 2018 ATP Finals championship Alexander Zverev, 21, Germany, World No 9 Kei Nishikori, 29, Japan, World No 12 Borna Coric, 22, Croatia, World No 13 Fabio Fognini, 31, Italy, and World No 19 Daniil Medvedev, 22 Russia, advanced to the second round of Australian Open (AO) 2019 at Melbourne Park, Melbourne, Australia, on January 15, 2019.

Zverev beat World No 67 Aljaz Bedene, 29, Slovenia, in straight sets 6-4, 6-1, 6-4 in the first round. Zverev fired 11 Aces in the match to 7 by Bedene. He, however, also committed 6 double faults to 2  by Bedene. Zverev won 6 break points to 2 by his opponent. Zverev won 8 games in a row in the match to 2 by Bedene.

Nishikori entered the second round after his opponent and World No 176 Kamil Majchrzak, 23, Poland, retired in the 5th set trailing 3-0. Nishikori had won first two sets 6-3, 7-6. But Kamil won the next two sets 6-0, 6-2 but retired in the 5th set due to injury. Nishikori won 8 break points in the match to 2 by Kamil. Kamil had fired 11 Aces in the match to 6 by Nishikori.

Coric beat World No 321 Steve Darcis, 34, Belgium, in straight sets 6-1, 6-4, 6-4 to enter the second round. Coric fired 16 Aces in the match to 11 by Darcis. Coric committed 3 double faults to 1 by Darcis. But, Coric won 4 break points besides winning 5 games in a row.

Fognini entered the second round after his opponent and World No 79 Jaume Munar, 21, Spain, retired in the third set 3-1 in their first round match. Fognini had won the first two sets 7-6, 7-6 and was leading 3-1 in the third.

Medvedev beat World No 119 Lloyd Harris, 21, South Africa, 6-1, 6-2, 6-1 in the first round. Medvedev served 7 Aces in the match to 6 by Harris. Both the players committed 4 double faults each in the match. Medvedev though won 7 break points in the match besides winning 5 games in a row.

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