Find all complex numbers such that
z^2=2i. Thank you!
it's easiest to use the polar form
if z = r cisθ
z^2 = r^2 cis2θ
so,
r^2 cis2θ = 2 cis π/2
r = ±√2
θ = π/4
So, we have
√2 cis π/4
-√2 cis π/4
√2/√2 + √2/√2 i = 1+i
or -(1+i)
check:
(1+i)^2 = 1+2i+i^2 = 1+2i-1 = 2i
same for -(1+i)
To find all complex numbers that satisfy the equation z^2 = 2i, we can use several methods. One approach is to express z in terms of its real and imaginary parts and substitute them into the equation.
Let z = a + bi, where a and b are real numbers representing the real and imaginary parts of z, respectively.
Substituting z into the equation, we have:
(a + bi)^2 = 2i
Expanding the square, we get:
a^2 + 2abi - b^2 = 2i
Equating the real and imaginary parts of both sides, we get:
a^2 - b^2 = 0 (the real part)
2ab = 2i (the imaginary part)
From the first equation, we have a^2 = b^2, which implies a = ±b.
Substituting a = ±b into the second equation, we obtain:
2(±b)b = 2i
2b^2 = 2i
b^2 = i
b = ±√i
To find the values of a corresponding to each value of b, we substitute b into the equation a = ±b:
a = ±(±√i)
Simplifying the expression, we have:
a = ±(±(1/√2 + 1/√2)i)
Combining the real and imaginary parts, we get four possible complex numbers that satisfy the equation z^2 = 2i:
z = ±(1/√2 + 1/√2)i, ±(1/√2 - 1/√2)i
Therefore, the complex numbers that satisfy z^2 = 2i are:
z = (1 + i)/√2, (-1 + i)/√2, (1 - i)/√2, (-1 - i)/√2