Jenson Brooksby keeps on rising. The American player reached the second round of Indian Wells after defeating Turkish qualifier Cem Ilkel in the first round 7-6 6-4. Jenson gained 10 places in the live rankings to be the new world no.69, which represents his career-high.
He will face Alexander Zverev in the second round. If he upsets the German player, Brooksby will be ranked either 65 or 66 in the live rankings.
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Brooksby in Indian Wells, ranking, and results in 2021
Currently the world no.76 (career-high), Jenson owns an overall 42-10 record in 2021. Jenson is now having a run in Indian Wells where he toppled
Jenson Brooksby will play the world no.4 Alexander Zverev in the 2nd round. They have never competed against each other till now in top tournaments.
Jenson clinched 3 titles in 2021 in the Potchefstroom 2 Challenger, in the Orlando Challenger and in the Tallahassee Challenger. The American was the runner-up in the Cleveland Challenger and in Newport (Hall of Fame Open).
Alexander Zverev in 2021
Currently ranked no.4, Zverev played his last match on the 25th of September when he overcame world no.23 John Isner 7-6(5) 6-7(6) 10-5 in the round robin in the Laver Cup (draw).
During this season Zverev has an overall 44-12 win-loss record. The German clinched 4 titles in 2021 in Acapulco, in Madrid, in Tokyo and in Cincinnati.
Jenson Brooksby and Alexander Zverev have never faced off.
Jenson Brooksby and Alexander Zverev |
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