Q. What is the solution if i^i?

• i=cos(pi/2)+i*sin(pi/2)=e^(i*pi/2)
  so i^i={e^(i*pi/2)}^i
  =e^(i*i*pi/2)
  =e^(i^2*pi/2)
  =e^(-pi/2)
  =cos(-pi/2)+i*sin(-pi/2)
  =0+i*(-1)
  =-i
• 11 years agoHelpfull: Yes(44) No(3)
• ANS: -i
  i=cos(90)+i*sin(90)=e^(i*90)
  so i^i=e^(i*90*i)=e^(-90i)=cos(-90)+i*sin(-90)=-i.
• 11 years agoHelpfull: Yes(6) No(0)
• i^i = e^(-pi/2)
  
• 11 years agoHelpfull: Yes(6) No(4)
• i^i=-1 i.e.,according to complex number analysis
• 11 years agoHelpfull: Yes(2) No(4)
• If iota, i = sqrt(-1)
  
  Then, i^i = sqrt(-1) x sqrt(-1)
  = -1
• 11 years agoHelpfull: Yes(1) No(2)
• i is root over -1. i sq is -1.
• 11 years agoHelpfull: Yes(1) No(0)
• The value of i^2 is -i
  
• 11 years agoHelpfull: Yes(1) No(1)
• i=cos(pi/2)+i*sin(pi/2)=e^(i*pi/2)
  now,
  i^i={e^(i*pi/2)}^i
  =e^(i^2*pi/2)
  =e^(-pi/2) [since i^2=-1]
  =cosh(pi/2)-sinh(pi/2)
  =0-1
  i^i=-1.
  WHICH IS REQUIRED VALUE OF i^i.
• 10 years agoHelpfull: Yes(1) No(0)
• A)
  cos@ + i.sin@ = e^(i.@)
  if we take @ = pi/2:
  i = e^(i.pi*/2)
  Raise both sides to exponent i:
  i^i = e^ (i.pi/2 . i)
  i^i = e^(-pi/2)
  
• 10 years agoHelpfull: Yes(1) No(0)
• value is -1
• 11 years agoHelpfull: Yes(0) No(1)
• Thanq u rakesh and ranjit.
  i am in favour of ur ans.
• 11 years agoHelpfull: Yes(0) No(1)