# Thread: Roots of an equation (complex numbers)

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

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

Another attempt : $z^2-2i=0 \Leftrightarrow a^2-b^2+i(2ab)=2i \Rightarrow a=b=1 \Rightarrow z=1+i$ (easy to graph).
I notice that $|z|=\sqrt 2$ 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 : $z^2-2i=0$ and graph them.
My attempt : $z^2-2i=0 \Leftrightarrow r^2e^{2i\theta}=2i \Rightarrow r^2=2$ and $e^{2i\theta}=i$ (I don't trust the implication here) and I don't reach anything.

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

So what are the roots? I didn't know an equation could have roots!
My attempt would be $z^2-2i=0 \Leftrightarrow z^2-(\sqrt{2}\times -1)^2=0 \Leftrightarrow (z-(\sqrt{2}\times -1))(z+(\sqrt{2}\times -1))$ $\Leftrightarrow (z+\sqrt{2})(z-\sqrt{2}) \Leftrightarrow z=\pm\sqrt{2}$

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

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

As a check, note that $(1 + i)^2 = 1 + 2i - 1 = 2i$.

If you take the polar approach, then you should note that

$e^{2 i \theta} = i \Rightarrow e^{2 i \theta} = e^{\pi/2 + 2 n \pi}$ where n is an integer

$\Rightarrow 2 \theta = \frac{\pi}{2} + 2 n \pi \Rightarrow \theta = \frac{\pi}{4} + n \pi$.

n = 0: $\theta = \frac{\pi}{4}$.

n = 1: $\theta = \frac{5 \pi}{4}$.

Therefore the two roots are:

$z = \sqrt{2} e^{i \pi/4} = \sqrt{2} \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 1 + i$

$z = \sqrt{2} e^{i 5 \pi/4} = \sqrt{2} \left( - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = -1 - i$

Originally Posted by pickslides
My attempt would be $z^2-2i=0 \Leftrightarrow z^2-(\sqrt{2}\times -1)^2=0 \Leftrightarrow (z-(\sqrt{2}\times -1))(z+(\sqrt{2}\times -1))$ $\Leftrightarrow (z+\sqrt{2})(z-\sqrt{2}) \Leftrightarrow z=\pm\sqrt{2}$
Unfortunately this is very wrong, as expanding shows. You've found the roots of $z^2 -2 = 0$.