# Thread: Roots of an equation (complex numbers)

1. ## Roots of an equation (complex numbers)

I must find the roots of the following equation : $\displaystyle z^2-2i=0$ and graph them.
My attempt : $\displaystyle z^2-2i=0 \Leftrightarrow r^2e^{2i\theta}=2i \Rightarrow r^2=2$ and $\displaystyle e^{2i\theta}=i$ (I don't trust the implication here) and I don't reach anything.

Another attempt : $\displaystyle z^2-2i=0 \Leftrightarrow a^2-b^2+i(2ab)=2i \Rightarrow a=b=1 \Rightarrow z=1+i$ (easy to graph).
I notice that $\displaystyle |z|=\sqrt 2$ in my both attempts.

So what are the roots? I didn't know an equation could have roots!

2. Originally Posted by arbolis
I must find the roots of the following equation : $\displaystyle z^2-2i=0$ and graph them.
My attempt : $\displaystyle z^2-2i=0 \Leftrightarrow r^2e^{2i\theta}=2i \Rightarrow r^2=2$ and $\displaystyle e^{2i\theta}=i$ (I don't trust the implication here) and I don't reach anything.

Another attempt : $\displaystyle z^2-2i=0 \Leftrightarrow a^2-b^2+i(2ab)=2i \Rightarrow a=b=1 \Rightarrow z=1+i$ (easy to graph).
I notice that $\displaystyle |z|=\sqrt 2$ in my both attempts.

So what are the roots? I didn't know an equation could have roots!
My attempt would be $\displaystyle z^2-2i=0 \Leftrightarrow z^2-(\sqrt{2}\times -1)^2=0 \Leftrightarrow (z-(\sqrt{2}\times -1))(z+(\sqrt{2}\times -1)) $$\displaystyle \Leftrightarrow (z+\sqrt{2})(z-\sqrt{2}) \Leftrightarrow z=\pm\sqrt{2} 3. Originally Posted by arbolis I must find the roots of the following equation : \displaystyle z^2-2i=0 and graph them. My attempt : \displaystyle z^2-2i=0 \Leftrightarrow r^2e^{2i\theta}=2i \Rightarrow r^2=2 and \displaystyle e^{2i\theta}=i (I don't trust the implication here) and I don't reach anything. Another attempt : \displaystyle z^2-2i=0 \Leftrightarrow a^2-b^2+i(2ab)=2i \Rightarrow a=b=1 \Rightarrow z=1+i (easy to graph). I notice that \displaystyle |z|=\sqrt 2 in my both attempts. So what are the roots? I didn't know an equation could have roots! The other solution to your cartesian approach is a = b = -1. So the roots are \displaystyle z = \pm (1 + i). As a check, note that \displaystyle (1 + i)^2 = 1 + 2i - 1 = 2i. If you take the polar approach, then you should note that \displaystyle e^{2 i \theta} = i \Rightarrow e^{2 i \theta} = e^{\pi/2 + 2 n \pi} where n is an integer \displaystyle \Rightarrow 2 \theta = \frac{\pi}{2} + 2 n \pi \Rightarrow \theta = \frac{\pi}{4} + n \pi. n = 0: \displaystyle \theta = \frac{\pi}{4}. n = 1: \displaystyle \theta = \frac{5 \pi}{4}. Therefore the two roots are: \displaystyle z = \sqrt{2} e^{i \pi/4} = \sqrt{2} \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 1 + i \displaystyle z = \sqrt{2} e^{i 5 \pi/4} = \sqrt{2} \left( - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = -1 - i Originally Posted by pickslides My attempt would be \displaystyle z^2-2i=0 \Leftrightarrow z^2-(\sqrt{2}\times -1)^2=0 \Leftrightarrow (z-(\sqrt{2}\times -1))(z+(\sqrt{2}\times -1))$$\displaystyle \Leftrightarrow (z+\sqrt{2})(z-\sqrt{2}) \Leftrightarrow z=\pm\sqrt{2}$
Unfortunately this is very wrong, as expanding shows. You've found the roots of $\displaystyle z^2 -2 = 0$.