I was solving practice problems for my upcoming midterm and however I got stuck with this question type.
It is asking me to find all roots and then sketch it.
$(1+i\sqrt{3})^{1/2}$
How do we proceed?
Asked
Active
Viewed 83 times
0
-
Try writing it in the form $re^{i\theta}$. – Jair Taylor Mar 23 '19 at 16:31
3 Answers
0
Hint:Substitute $$\sqrt{1+\sqrt{3}i}=a+bi$$ then you have to solve $$a^2-b^2=1$$ and $$\sqrt{3}=2ab$$
Dr. Sonnhard Graubner
- 95,283
0
$1+i\sqrt{3}=2e^{i\frac{\pi}{3}}={(\sqrt{2}e^{i\frac{\pi}{6}})}^2$ thus the roots are $\pm \sqrt{2}e^{i\frac{\pi}{6}}=\pm(\frac{\sqrt{6}}{2}+i\frac{\sqrt{2}}{2})$
General way to find the roots of $z=a+ib$
Let $\delta=x+iy$ a root of $z$. Then $\delta^2=x^2-y^2+2ixy=a+ib$ and thus $x^2-y^2=a$ and $2xy=b$. Using modulus we obtain : $|z|=|\delta|^2\Longrightarrow x^2+y^2=\sqrt{a^2+b^2}$.
This provides the following system: $$\begin{cases}x^2-y^2=a ,\\x^2+y^2=\sqrt{a^2+b^2},\\2xy=b \end{cases}.$$
The first two equations gives $$x^2=\frac{a+\sqrt{a^2+b^2}}{2}$$ $$ y^2=\frac{a-\sqrt{a^2+b^2}}{2}.$$
$$x=\pm\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}, y=\pm\sqrt{\frac{a-\sqrt{a^2+b^2}}{2}}.$$
The last equation can be used to find the sign of $x$ and $y$. In your example $2xy=\sqrt{3}>0$ which means that $sign(x)=sign(y)$ and thus we have to choose $x>0$ and $y>0$, or $x<0$ and $y<0$.
DINEDINE
- 6,081
0
Let the roots be of the form $u+iv$. You need to solve
$$(u+iv)^2=1+i\sqrt3.$$
As $(u+iv)^2=u^2-v^2+2iuv$, by identification
$$\begin{cases}u^2-v^2=1,\\2uv=\sqrt3\end{cases}.$$
Now multiplying the first by $u^2$,
$$u^4-u^2v^2=u^4-\frac34=u^2$$ is a biquadratic equation. We obtain
$$u^2=\frac{1\pm\sqrt{1+3}}{2}=\frac32,-\frac12$$
and the only real roots are
$$u=\pm\frac{\sqrt3}{\sqrt2},$$ corresponding to $$v=\pm\frac1{\sqrt2}.$$
Final check:
$$\left(\frac{\sqrt3}{\sqrt2}+\frac i{\sqrt2}\right)^2=\frac32-\frac12+2i\frac{\sqrt3}{2}$$