Express the following in the form a + bi, where a and b are real numbers:
√i
i = (1,π/2)
so, √i = (1,π/4) = 1/√2 + 1/√2 i
To express √i in the form a + bi, where a and b are real numbers, we need to find values for a and b that satisfy the equation √i = a + bi.
Let's first represent i as a complex number: i = 0 + 1i.
Now, let's assume √i = a + bi, where a and b are real numbers.
We can square both sides of this equation to eliminate the square root symbol:
(√i)^2 = (a + bi)^2
Simplifying,
i = a^2 + 2abi - b^2
We can equate the real and imaginary parts of both sides of the equation:
For the real parts: 0 = a^2 - b^2
For the imaginary parts: 1 = 2ab
From the first equation, we have a^2 - b^2 = 0. This can be factored as (a + b)(a - b) = 0.
So, we have two possibilities to consider:
1) a + b = 0
Solving this equation, we get a = -b.
Substituting this value of a in the second equation, we have 1 = -2b^2.
Rearranging the equation, b^2 = -1/2, which has no real solutions.
2) a - b = 0
Solving this equation, we get a = b.
Substituting this value of a in the second equation, we have 1 = 2a^2.
Rearranging the equation, a^2 = 1/2, which has two possible solutions: a = √(1/2) or a = -√(1/2).
Since we are looking for real numbers a and b, we can take the positive value for a.
Therefore, a = √(1/2) and b = √(1/2).
Therefore, √i = √(1/2) + √(1/2)i.
Hence, √i can be expressed in the form a + bi as √(1/2) + √(1/2)i.