Compute the possible square roots

Question

compute the square root of 1 + i√3

I'm not sure if i am approaching this right but would it be √1+ √i√√3. then this square root of 1 equals 1, the square root of i equals -1, and then square root of three being squared again just equals 3. so then the answer is -2.

Answer 1

Solve

$$z^2=1+i\sqrt3.$$

With $z=a+ib$,

$$a^2-b^2+i2ab=1+i\sqrt3.$$

Then after identification of the real and imaginary parts

$$(a^2-b^2)^2+4a^2b^2=1^2+(\sqrt3)^2=2^2=(a^2+b^2)^2$$

so that ($a^2+b^2$ cannot be negative),

$$a^2-b^2+a^2+b^2=2a^2=1+2.$$

The solutions are

$$a=\pm\sqrt{\frac32}$$ with

$$b=\pm\frac1{\sqrt2}.$$

Answer 2

You are overcomplicating things. It's as easy as finding all possible combinations of $a$ and $b$ such that $(a+bi)^2=1+i\sqrt{3}$

Answer 3

Well, there's a lot going on here. First, you're mixing up "squaring" with "square root." For instance, the square root of $\sqrt{3}$ isn't $3$, rather $\sqrt{\sqrt{3}}$.

Second, square-rooting doesn't distribute over addition. You can't write this:

$$5 = \sqrt{25} = \sqrt{9+16} =\sqrt{9}+\sqrt{16} = 3 + 4 = 7$$

because 5 does not equal 7. The error occurs at the third equal sign where we split the square root into two piece. So that's a no-no.

Instead, I bet you're supposed to put the complex number in "polar form." If you can do that, you're more than half-way there.