x^2 = i
x = ±√i
Now, i = (1,π/2)
so x = ±(1,π/4)
or, (1,π/4), (1,5π/4)
Find all solutions and leave answers in trigonometric form.
x^2 - i = 0
Please help! I have no idea how to solve this!
2 answers
let x = a + bi , where a and b are real numbers
x^2 = a^2 + 2abi + b^2i^2, but i^2 = -1
= a^2 - b^2 + 2abi
= i <----- that was given
by comparison, a^2 - b^2 = 0
a^2 = b^2
a = ± b
and 2abi = i
2ab = 1
if a= +b, 2b^2 = 1
b = ± 1/√2 , and a = ± 1/√2 or ± √2/2
if a = -b, 2b(-b) = 1
b^2 = -1/2, no solution for real values of b
so x = a + bi
= (√2/2 + √2/2i) OR (-√2/2 + √2/2i)
= 1(cos45° + i cos45°) or 1(-cos225° + i sin225°)
= ± cosπ/4 + i sinπ/4
check one of them
x = √2/2 + √2/2i
x^2 = (√2/2 + √2/2i)^2
= 1/2 + 2(1/2) i + (1/2) i^2
= 1/2 + i - 1/2
= i
or, using De Moivre's theorem
x^2 = 1^2 (cos 2(45°) + i sin 2(45°) )
= cos90 + isin90
= 0 + 1i = i
x^2 = a^2 + 2abi + b^2i^2, but i^2 = -1
= a^2 - b^2 + 2abi
= i <----- that was given
by comparison, a^2 - b^2 = 0
a^2 = b^2
a = ± b
and 2abi = i
2ab = 1
if a= +b, 2b^2 = 1
b = ± 1/√2 , and a = ± 1/√2 or ± √2/2
if a = -b, 2b(-b) = 1
b^2 = -1/2, no solution for real values of b
so x = a + bi
= (√2/2 + √2/2i) OR (-√2/2 + √2/2i)
= 1(cos45° + i cos45°) or 1(-cos225° + i sin225°)
= ± cosπ/4 + i sinπ/4
check one of them
x = √2/2 + √2/2i
x^2 = (√2/2 + √2/2i)^2
= 1/2 + 2(1/2) i + (1/2) i^2
= 1/2 + i - 1/2
= i
or, using De Moivre's theorem
x^2 = 1^2 (cos 2(45°) + i sin 2(45°) )
= cos90 + isin90
= 0 + 1i = i
3 answers