Complex numbers question?

4 ⤊

Complex numbers question?

Solve the equation z² - 2iz -5 = 0, giving your answer in the form x + iy where x and y are real.

2007-10-25 23:03:13 · 4 answers · asked by plolol 2 in Science & Mathematics ➔ Mathematics

4 answers

z^2 - 2iz - 5 = 0

Using the quadratic formula :
z = [2i ± sqrt(4i^2 - 4(1)(-5)] / 2

= [2i ± sqrt(16)] / 2

= (2i ± 4) / 2

= i ± 2

In x + iy form, solutions are : 2 + i1 or -2 + i1

2007-10-26 00:22:40 · answer #1 · answered by falzoon 7 · 1⤊ 0⤋

z= ((2i+-sqrt(-4+20))/2 = +-2+i

2007-10-26 00:25:30 · answer #2 · answered by santmann2002 7 · 1⤊ 0⤋

2iz

2017-01-19 11:42:20 · answer #3 · answered by ? 4 · 0⤊ 0⤋

is a calulator any help?!

2007-10-25 23:06:25 · answer #4 · answered by Anonymous · 2⤊ 0⤋