Find the square root of the complex number z = 1 - 2i .

To find the square root of the complex number z = 1 - 2i, we’ll use the formula for finding the square root of a complex number.

Let z = a + bi be a complex number, where a and b are real numbers.

The square root of z is given by:

\$ sqrt(z) = \pm(sqrt(( sqrt(a^2 + b^2) + a)/2) + isign(b) * sqrt((sqrt(a^2 + b^2) - a)/2)) \$

Where sqrt represents the square root operation, and sign(b) is the sign function that returns -1 for b < 0, 0 for b = 0, and 1 for b > 0.

Now, let’s find the square root of the complex number z = 1 - 2i:

a = 1, b = - 2

Calculate the magnitude of the complex number

\$ | r | = sqrt(a^2 + b^2) = sqrt(1^2 + (-2)^2) = sqrt(1 + 4) = sqrt(5) \$

Plugg the values in the formula:

\$ sqrt(z) = \pm(sqrt((sqrt(5) + 1)/2) + isign(-2) * sqrt((sqrt(5) - 1)/2) )\$

After simplification

\$ sqrt(z) = \pm(sqrt((sqrt(5) + 1)/2) - isqrt((sqrt(5) - 1)/2))\$

So, the square root of the complex number z = 1 - 2i is:

\$ sqrt(z) = \pm(sqrt((sqrt(5) + 1)/2) - isqrt((sqrt(5) - 1)/2))\$

This option provides accurate representations for the square root of the specified complex number

\$ sqrt(z) = \pm(sqrt((sqrt(5) + 1)/2) - isqrt((sqrt(5) - 1)/2))\$

It is wrong ecause it doesn’t follow the correct formula for the square root of a complex number

\$ \sqrtz = ± (\sqrt((2 ± \sqrt5)/2) + i\sqrt((1 ± \sqrt5)/2)) \$

Option C differs from the correct answer due to a sign difference in the imaginary part

\$ sqrt(z) = \pm(sqrt((sqrt(5) + 3)/2) - isqrt((sqrt(5) - 3)/2))\$

TThis option has incorrect expressions for the square root of the given complex number

\$ sqrt(z) = \pm(sqrt((sqrt(2) + 3)/2) + isqrt((sqrt(2) - 3)/2))\$

Option

A

Can you summarize what you’ve understood in the above steps?