f(z)=z2+c
- #924,421
- Sep 17, 2014 6:10am Sep 17, 2014 6:10am
- Joined Dec 2013 | Status: Trader | 2,141 Posts
f(z)=z2+c
- #924,425
- Sep 17, 2014 6:31am Sep 17, 2014 6:31am
- Joined Apr 2013 | Status: I'm learnding! | 8,973 Posts
- #924,426
- Sep 17, 2014 6:47am Sep 17, 2014 6:47am
- Joined Dec 2010 | Status: Foook Bollinger-dr.Kegel knows! | 9,794 Posts
- #924,428
- Sep 17, 2014 6:50am Sep 17, 2014 6:50am
- Joined Apr 2013 | Status: I'm learnding! | 8,973 Posts
- #924,431
- Sep 17, 2014 7:00am Sep 17, 2014 7:00am
- Joined Dec 2013 | Status: Trader | 2,141 Posts
f(z)=z2+c
- #924,436
- Sep 17, 2014 7:12am Sep 17, 2014 7:12am
- Joined Dec 2010 | Status: Foook Bollinger-dr.Kegel knows! | 9,794 Posts
- #924,437
- Sep 17, 2014 7:15am Sep 17, 2014 7:15am
- Joined May 2014 | Status: From1toMillion | 9,468 Posts
Everyone can see the chart, but only a few can actually read it.
- #924,440
- Sep 17, 2014 7:23am Sep 17, 2014 7:23am
- Joined May 2014 | Status: From1toMillion | 9,468 Posts
Everyone can see the chart, but only a few can actually read it.