Q407) Complex Numbers?

Question

Given that z = x + iy where x, y are real,

i) find z^2 in terms of x and y and
ii) find both square roots of i.

Answer 1

i)
z^2 = x^2 - y^2 + 2ixy

ii)
(1+i)/sqrt2

-(1+i)/sqrt2

i=cis(pi/2)

in general i = cis(2npi + (pi/2))

i ^(1/2) = cis[ (2npi + (pi/2))/2 ] n = 0 , 1

Answer 2

(x+i*y)^2=x^2+y^2-2xyi
let i^1/2=a+bi
=>i=a^2-b^2+2abi
=>a^2=b^2 & 2ab=1
=>a=+b or a=-b
& ab=1/2
=>a^2=1/2 =>a=1/(2)^1/2 or a = -1/(2)^1/2
& a=1/i(2)^1/2 or a=-1/i(2)^1/2

Answer 3

i)
z² = (x + iy).(x + iy)
z² = x² + 2i.xy - y²
z² = (x² - y²) + 2i.xy

ii)
i² = - 1
i = ± √(-1)