z=3+4i use the algebraic method to find the square root of z

To find the square root of z, we need to find a number w such that w^2 = z.

Let's assume w = a + bi, where a and b are real numbers.

Now, we can square w: w^2 = (a + bi)^2 = (a + bi)(a + bi) = a^2 + 2abi + b^2(i^2) = a^2 + 2abi + b^2(-1) = a^2 + 2abi - b^2

Since z = 3 + 4i, we can equate the real and imaginary parts of w^2 and z: a^2 - b^2 = 3 ...(1) 2ab = 4 ...(2)

From equation (2), we can solve for a: a = 2/b

Substituting this value of a into equation (1): (2/b)^2 - b^2 = 3 4/b^2 - b^2 = 3 4 - b^4 = 3b^2 b^4 + 3b^2 - 4 = 0

Now, let's solve this quadratic equation for b^2: (b^2 - 1)(b^2 + 4) = 0

Setting each factor equal to zero: b^2 - 1 = 0 or b^2 + 4 = 0

For b^2 - 1 = 0: b^2 = 1 b = ±1

For b^2 + 4 = 0: b^2 = -4 b = ±√(-4) b = ±2i

Therefore, we have four possible values for b: b = 1, b = -1, b = 2i, or b = -2i.

Using the value of b, we can find the corresponding values of a using a = 2/b:

For b = 1: a = 2/1 = 2

For b = -1: a = 2/(-1) = -2

For b = 2i: a = 2/(2i) = 1/i = -i

For b = -2i: a = 2/(-2i) = -1/i = i

So, the four possible values of w are: w1 = 2 + i w2 = -2 - i w3 = -i w4 = i

Therefore, the square roots of z = 3 + 4i are: √z1 = 2 + i √z2 = -2 - i √z3 = -i √z4 =