Former US Open champion Juan Martin del Potro eased into the second round of the Indian Wells WTA tournament with a 6-4, 6-0 demolition of Czech Radek Stepanek on Thursday.
Argentine del Potro, who played just three tournaments last year because of an injury to his right wrist, broke his opponent twice in the opening set and three times in the second to seal victory in 84 minutes.
"I broke his serve very soon and took confidence with my serve," the six-foot six-inch del Potro told reporters.
"That helped my whole game, and with my forehand especially. I played a good match," he added.
It was the first time the Argentine had beaten Stepanek in three meetings, having lost to the Czech twice in 2009.
"It was very close in those matches but today I played better than him," said del Potro, who won his eighth ATP title at last month's Delray Beach International Championships.
"I made the most of my opportunities, especially in the second set. I was very focused on my serve, trying to win very fast the games," he added.
Del Potro, who has slipped to 90th in the rankings from a career-high fourth, will next meet Croat Ivan Ljubicic, the tournament's 14th seed who beat American Andy Roddick 7-6, 7-6 in last year's final at Indian Wells.
"He has a very strong serve and a nice backhand but I know if I play good tennis, if I focus the whole match, I have a chance to beat him," del Potro said of Ljubicic.
"But it will be very tough for both of us," he added.
In other men's matches on Thursday, Serb Janko Tipsarevic crushed Germany's Tobias Kamke 6-2, 6-0 while American qualifier Donald Young came from a set down to beat Italian Potito Starace 2-6, 6-3, 6-3.
In the women's tournament, Poland's Urszula Radwanska scraped past Serb Bojana Jovanovski 3-6, 6-4, 7-6 and Czech qualifier Lucie Hradecka beat Belgium's Kirsten Flipkens 7-6, 6-2.
The seeds swing into action on Friday when Belgian world number two Kim Clijsters takes on Russian Alla Kudryavtseva and defending champion Jelena Jankovic of Serbia faces American wildcard Coco Vandeweghe.
