Hi everybody,
Find the square roots of the following in cartesian form.
$\displaystyle 1+sqrt3i$
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.
You have to find two real numbers a and b such as $\displaystyle (a+bi)^2=1+i\sqrt{3}$
Square the left side member and we get $\displaystyle a^2-b^2+2abi=1+i\sqrt{3}$. Then
$\displaystyle \left\{\begin{array}{ll}a^2-b^2=1\\2ab=\sqrt{3}\end{array}\right.$
You can add another equation to the system.
$\displaystyle |(a+bi)^2|=|1+i\sqrt{3}|\Rightarrow a^2+b^2=2$.
Then the system is
$\displaystyle \left\{\begin{array}{ll}a^2-b^2=1\\2ab=\sqrt{3}\\a^2+b^2=2\end{array}\right.$
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.
This is the standard way.
Because $\displaystyle 1 + i\sqrt 3 = 2\left( {\cos \left( {\frac{\pi }{3}} \right) + i\sin \left( {\frac{\pi }{3}} \right)} \right)$ the two square roots are:
$\displaystyle \sqrt 2 \left( {\cos \left( {\frac{\pi }
{6}} \right) + i\sin \left( {\frac{\pi }
{6}} \right)} \right)\;\& \,\sqrt 2 \left( {\cos \left( {\frac{{ - 5\pi }}
{6}} \right) + i\sin \left( {\frac{{ - 5\pi }}
{6}} \right)} \right)$