Question

How do you find the square roots of `-25i`?

Answer 1

The two square roots, which are negatives of each other, are:

`({5\sqrt{2}}/2)-({5\sqrt{2}}/2)i`

`(-{5\sqrt{2}}/2)+(\{5sqrt{2}}/2)i`.

Explanation:

Here is how to find the square roots of any conplex number `a+bi` with `b` nonzero.

We seek `x+yi` such that

`(x+yi)^2=a+bi`.

Remember the product rule for complex numbers, thus

`(x+yi)^2=(x^2-y^2)+2xyi=a+bi`.

Match real and imaginary parts:

`x^2-y^2=a` Equation 1
`2xy=b` Equation 2.

We have another relation, the magnitude of `(x+yi)^2=a+bi` is the square of the magnitude of `x+yi`:

`x^2+y^2=\sqrt{a^2+b^2}` Equation 3

Now take the average of Equations 1 and 3:

`x^2={\sqrt{a^2+b^2}+a}/2`

`color(blue)(x=\pm\sqrt{{\sqrt{a^2+b^2}+a}/2})`

Now take half the difference between Equations 1 and 3:

`y^2={\sqrt{a^2+b^2}-a}/2`

`color(blue)(y=\pm\sqrt{{\sqrt{a^2+b^2}-a}/2})`

One more thing remains: choose the right signs. That is where Equation 2 comes in.

`color(blue)("x and y have the same sign if b is positive")`

`color(blue)("x and y have opposite sign if b is negative")`

Now put this all together for the square root of `-25i`.

`a=0, b=-25`.

`b` is negative so choose opposite signs for `x` and `y`.

`\sqrt{a^2+b^2}=25`.

`x=\pm\sqrt{{\sqrt{a^2+b^2}+a}/2}=\pm\sqrt{(25+0)/2}=\pm{5\sqrt{2}}/2`.

`y=\pm\sqrt{{\sqrt{a^2+b^2}-a}/2}=\pm\sqrt{(25-0)/2}=\pm{5\sqrt{2}}/2`.

So remembering that we have opposite signs for this case we have the two square roots

`({5\sqrt{2}}/2)-({5\sqrt{2}}/2)i`

`({-5\sqrt{2}}/2)+({5\sqrt{2}}/2)i`