Find the square roots of 3+4i. The answer should be in the form a+bi?
The answer that i got was 2+i, but it did not seem to work. Any advice on what might be happening in this problem, and how to solve it?
well, what is (2+i)^2 ?
2^2 + 4i + i^2 = 3+4i
Looks like you got it right.
But could it also have been -2 - i ?
since (-2-i)^2 = 3+4i
To find the square roots of a complex number, such as 3+4i, you can use the following steps:
Step 1: Express the complex number in the form a+bi. In this case, 3+4i is already in the desired form.
Step 2: Let's assume the square root of 3+4i is of the form (a+bi). Therefore, (a+bi)^2 = 3+4i.
Step 3: Expanding (a+bi)^2, we get a^2 + 2abi - b^2 = 3+4i.
Step 4: Comparing the real and imaginary parts, we have two equations:
Real part: a^2 - b^2 = 3 ------ (Equation 1)
Imaginary part: 2ab = 4 ------ (Equation 2)
Step 5: Solve Equation 2 for a:
2ab = 4 => ab = 2 ------ (Equation 3)
We can divide Equation 3 by b, assuming b is not zero (otherwise, it would result in a divide-by-zero error):
a = 2/b ------ (Equation 4)
Step 6: Substitute Equation 4 into Equation 1 to solve for b:
(2/b)^2 - b^2 = 3
4/b^2 - b^2 = 3
4 - b^4 = 3b^2
b^4 + 3b^2 - 4 = 0
This equation is a quadratic equation in terms of b^2. Let's substitute b^2 with x and solve for x:
x^2 + 3x - 4 = 0
By factoring, we get:
(x + 4)(x - 1) = 0
So, x = -4 or x = 1.
Step 7: Substitute x back with b^2:
When x = -4: b^2 = -4 --> b = ±2i
When x = 1: b^2 = 1 --> b = ±1
Step 8: Substitute the values of b into Equation 4 to solve for a:
When b = 2i: a = 2/(2i) = -i
When b = -2i: a = 2/(-2i) = i
When b = 1: a = 2/(1) = 2
When b = -1: a = 2/(-1) = -2
Therefore, the square roots of 3+4i are:
(a+bi) = (-i+2i) = 2i and (a+bi) = (i-2i) = -i.
The answer you mentioned, 2+i, is incorrect. The correct roots are 2i and -i.