# Math Help - Complex Nuumbers

1. ## Complex Nuumbers

Hi everybody,

Find the square roots of the following in cartesian form.
$1+sqrt3i$
I think thats typed right.
I am not sure how to attempt it.
The example in my book is too simple and doesn't explain it properly.
Thanks.

2. You have to find two real numbers a and b such as $(a+bi)^2=1+i\sqrt{3}$

Square the left side member and we get $a^2-b^2+2abi=1+i\sqrt{3}$. Then

$\left\{\begin{array}{ll}a^2-b^2=1\\2ab=\sqrt{3}\end{array}\right.$

You can add another equation to the system.

$|(a+bi)^2|=|1+i\sqrt{3}|\Rightarrow a^2+b^2=2$.

Then the system is

$\left\{\begin{array}{ll}a^2-b^2=1\\2ab=\sqrt{3}\\a^2+b^2=2\end{array}\right.$

Find a and b from the first and the third equations. Then choose the signs for a and b such as the second equation is satisfyed.

3. ## Huh

Sorry but I haven't learnt any of that but I don't get it.
Can we have a simpler way. I know the eqaution a+root(a^2+b^2)/2
That is the only thing I know.
Thanks.

4. Originally Posted by Awsom Guy
Sorry but I haven't learnt any of that but I don't get it. Can we have a simpler way.
This is the standard way.
Because $1 + i\sqrt 3 = 2\left( {\cos \left( {\frac{\pi }{3}} \right) + i\sin \left( {\frac{\pi }{3}} \right)} \right)$ the two square roots are:
$\sqrt 2 \left( {\cos \left( {\frac{\pi }
{6}} \right) + i\sin \left( {\frac{\pi }
{6}} \right)} \right)\;\& \,\sqrt 2 \left( {\cos \left( {\frac{{ - 5\pi }}
{6}} \right) + i\sin \left( {\frac{{ - 5\pi }}
{6}} \right)} \right)$

5. What about the formula a+sqrt(a^2+b^2)/2.
I am sopposed to find x and then y. Then bring them together to make a cartesian formed equation like 2+i and -2-i.
Thanks.

6. If you analyze the solution by red_dog, you will find what you need.