can anyone solve and tell me the value of squareroot of -1
2007-03-29 23:45:46 · 8 answers · asked by legolas g/Frederich 4 in Science & Mathematics ➔ Mathematics
This is where you delve into complex numbers.
sqrt(-1) = i in mathematics, or j in engineering.
2007-03-30 02:32:37 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Negative numbers do not have a square root .
So the square root of -1 does not exist.
-1*-1 =1
Symbols such as i and j which are used in complex numbers are operators which turn a vector through 90 degrees anti clockwise. so if you see the term j^2 = -1 it means the vector has been turned through 180 deg . and j or i is not the square root of -1.
2007-03-30 03:33:16 · answer #2 · answered by mad_jim 3 · 0⤊ 1⤋
the imaginary unit i. Also denoted in electrical engineering by j as to avoid confusion with current.
It is defined that i^2 = -1
2007-03-30 01:41:55 · answer #3 · answered by SS4 7 · 1⤊ 0⤋
its undefined unless you use complex numbers to slove. take out common factor of j= -1 and solve
2007-03-30 00:03:57 · answer #4 · answered by katt2305 1 · 0⤊ 0⤋
Infinity
2007-03-29 23:48:52 · answer #5 · answered by gav 4 · 0⤊ 3⤋
"i" the imaginary number
seriously
also (sqroot)-9 = 3*i or 3i
2007-03-29 23:48:30 · answer #6 · answered by Justin H 4 · 1⤊ 0⤋
â - 1= no solution for real numbers
- - - - - - - -
Imiginary Number solution
â - 1=
i â 1 =
1i
- - - - - - s-
2007-03-30 00:20:43 · answer #7 · answered by SAMUEL D 7 · 0⤊ 0⤋
-1=1(cos(pi)+isin(pi))
zk=1^(1/2)
(cos(pi/n+2kpi/n)
+isin(pi/n+2kpi/n))
where zk is a root,k=0,1
and n=2
z0=1*(cos(pi/2+0)+isin(pi/2+0))
=0+i1=i
z1=1*(cos(pi/2+pi)
+isin(pi/2+pi))
=cos(3pi/2)+isin(3pi/2)
=0+i(-1)= -i
hence,sqrt(-1)=+i or-i
i hope that this helps
2007-03-30 02:06:21 · answer #8 · answered by Anonymous · 0⤊ 1⤋