> Notice that a^b=e^(bln(a)). Using this, compute both 2^i and 2^(i+1)?

Notice that a^b=e^(bln(a)). Using this, compute both 2^i and 2^(i+1)?

Notice that a^b=e^(bln(a)). Using this, compute both 2^i and 2^(i+1)?

Posted at： 2014-12-18　
2^i = e^(i ln 2) = cos (ln 2) + i sin (ln 2)

2^(i + 1) = 2^i * 2 = 2 cos (ln 2) + 2i sin (ln 2)

If you'd have put this in the mathamatics area, you'd have got an answer sooner.
  • Previous： What size should images be for websites (file size)? I've had lots of different answers=200kb/70kb/12kb?
  • Next：Creating a Binary Calculator in C#?