Find all complex numbers z such that z^2=2i. Write your solutions in form, separated by commas. So, "1+2i, 3-i" is an acceptable answer format, but "2i+1; -i+3" is not. Thank you!
-8+i
To find all complex numbers z such that z^2 = 2i, we can start by expressing z in terms of its real and imaginary parts. Let z = a + bi, where a and b are real numbers.
Substituting this into the equation z^2 = 2i, we get (a + bi)^2 = 2i.
Expanding the square, we have (a + bi)(a + bi) = 2i.
Using the distributive property, we have a^2 + 2abi - b^2 = 2i.
Separating the real and imaginary parts, we get:
a^2 - b^2 = 0 (1)
2ab = 2i (2)
From equation (1), we can solve for a^2 = b^2, which implies a = b or a = -b.
If we consider a = b, then substituting this value into equation (2), we get:
2a^2 = 2i.
This gives us a^2 = i. Taking the square root of both sides, we have a = ±√i.
To express √i in terms of a + bi, we can use the fact that (a + bi)^2 = 2i.
Expanding this equation, we have a^2 + 2abi + b^2i^2 = 2i.
Simplifying, we get a^2 - b^2 = 0 (equivalent to equation (1) above) and 2ab = 2.
From equation (1), we know that a = ±b, so substituting this into the equation 2ab = 2, we get 2a^2 = 2.
This gives us a^2 = 1, which implies a = ±1.
If a = 1, then b = ±1. Therefore, the solutions for z when a = b (which is a = b = ±1) are:
1 + i and -1 - i.
If a = -1, then b = ±1. Therefore, the solutions for z when a = -b (which is a = -b = ±1) are:
-1 + i and 1 - i.
Let's summarize the solutions:
z = 1 + i, -1 - i, -1 + i, and 1 - i.
So, the complex numbers z such that z^2 = 2i are 1 + i, -1 - i, -1 + i, and 1 - i.