Solve the equation z^2 - 2iz - 5 = 0, giving your answers in the form x + iy where x and y are real.
2007-07-10 11:53:33 · 6 answers · asked by plolol 2 in Science & Mathematics ➔ Mathematics
You can either plug it into the quadratic equation:
http://en.wikipedia.org/wiki/Quadratic_equation
or you can factor it:
(z - 2 - i) (z + 2 - i)
You can check that if you want, by multiplying it out. You get:
z² -2z - i z +2z - iz - (2+i)(2-i)
= z² - 2iz - 5
The quadratic equation is also fairly simple in this case, and I presume that not everyone is as good at complex-integer factoring as I am. It's straightforward:
(2i ± √(-4 - 4*(-5)) / 2
= (2i ± √(16) ) / 2
= i ± 2
The roots are:
2+i, -2+i
2007-07-10 12:04:27 · answer #1 · answered by сhееsеr1 7 · 0⤊ 1⤋
1+1=11
2007-07-10 18:56:50 · answer #2 · answered by Anonymous · 0⤊ 4⤋
You could use the quadratic formula for that.
z = [2i +/- sqrt(-4 + 20) ] / 2
= i +/- 2
SOlutions:
-2 + i
2 + i
2007-07-10 19:01:12 · answer #3 · answered by Dr D 7 · 0⤊ 1⤋
z = [2i +/- sqrt(-4 +20)]/2
z = (2i +/-4)/2
z = i+2 and z= i-2
2007-07-10 19:01:59 · answer #4 · answered by ironduke8159 7 · 0⤊ 1⤋
±2+i.
simply apply quadratic formula:
(-b±â[b²-4ac]) / 2a
{2i ±â(-4+20) } / 2 = i ± 2.
check: sum: 2i , and product: -5.
d:
2007-07-10 19:01:02 · answer #5 · answered by Alam Ko Iyan 7 · 0⤊ 2⤋
(0 + 3i) and (-2 + 0i)
2007-07-10 18:57:54 · answer #6 · answered by Smoove Davey P 2 · 0⤊ 2⤋