Find the two square roots of $i$

Question

I have this question I am stumped upon for my Test-Review:

Write $i$ as a complex number in polar form. Use the result and DeMoivre's Theorem to find the square roots of $i$.

I got the first question:

$i = a + bi$

$0 + 1i$

$z = \sqrt{0 ^ 2 + 1 ^ 2}$

$z = 1$

$\Theta = \tan^-1(y/x)$

$\Theta = \tan^-1(0/1)$

$\Theta = 90^\circ$

Therefore... answer is $1 \cdot cis(90)$... Ok I got that but the Demoivre theorem and the sqrt I couldn't understand... This is the answer.. from anwser key:

a) $cis(90^\circ)$

b) $cis(45 ^\circ) , circ(225 ^\circ)$

I'm following

My teacher follows same way as this: http://hotmath.com/hotmath_help/topics/polar-form-of-a-complex-number.html

Formulas Available

Answer 1

Okay, I have tried to do my best earlier... but go through this step by step.

$z_1 = a + i\,b\\ z_2 = x + i\,y \\ z_1z_2 = (a + i\,b)(x + i\,y) = (ax + i\,bx + i\,ay + i^2\,by)$

$i^2 = -1$, and you can combine the $i$ terms.

$z_1z_2 = (ax - by) + i\,(ay + bx)$

Okay, if you are already lost re-read this.

If $z_1$ and $z_2$ are in polar...

$z_1 = \rho_1 (\cos \theta + i \sin\theta)\\ z_2 = \rho_2 (\cos \phi + i \sin\phi)\\ z_1z_2 =\rho_1\rho_2 (\cos\theta\cos\phi - \sin\theta \sin\phi + i(\sin\theta \cos \phi + \cos \theta \sin \phi)$

This is the same multiplication that we did in rectangular coordinates. But notice how this ties out to the angle addition formulas for $\cos$ and $\sin$

$z_1z_2 =\rho_1\rho_2 (\cos(\theta+\phi) + i\sin(\theta+\phi))$

When we multiply complex numbers we add the angles! This is particularly noticeable in polar form, but it is just as true in Cartesian form.

$z^2 = z\cdot z =$$ \rho (\cos \theta + i \sin\theta)\rho (\cos \theta + i \sin\theta)\\ \rho^2 (\cos 2\theta + i \sin2\theta)\\$

De Moivre's theorem:

$z^n = \rho^n (\cos n\theta + i \sin n\theta)$

and it works just as well when $n$ is a fraction as it does when $n$ is a whole number.

$\sqrt z = z^{1/2}$

$\sqrt z = \pm \rho^{1/2} (\cos \frac{\theta}{2} + i \sin \frac{\theta}2)$

Everything up to this point is actually on your cheat-sheet in condensed from. But rather than just memorizing formula, try to understand what is actually going on, and where those formula are coming from.

And to the problem at hand:

$z = 0 + i\\ z = (\cos 90 + i \sin 90)\\ z^{1/2} = \pm (\cos 45 + i \sin 45)\\ z^{1/2} = (\cos 45 + i \sin 45), (\cos 225 + i \sin 225)$

Advanced stuff: This is not something you are expected to know in precaluculs, but if you ask the question on this board people may assume you know this.

Complex numbers in polar form can be translated into exponential form as follows:

$z = \rho e^{i\theta}\\ z^n = \rho^n e^{i\theta n}\\ z^{1/n} = \rho^\frac1n e^{\frac{i\theta}{n}}\\ $

Answer 2

You have $i = e^{i\pi/2}$ so this has two square roots, $e^{i\pi/4}$ and $-e^{i\pi/4} = e^{5i\pi/4}.$

Answer 3

Solution $1$

Let $z=x+iy\in \Bbb C$ such that $z^2=i$. then $z^2=x^2-y^2+2xyi=i$. It follows that

$$ x^2=y^2\iff x=\pm y\\ 2xyi=i \iff x=\frac 1 {2y} $$

Then, $x=\pm\frac 1 {2x}$, we get that $x=\pm\sqrt2/2$ , so we get $z_1=\sqrt 2/2+\sqrt 2 /2i$, $z_2=-z_1$.

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Having the $x,y$ coordinates, you can put these numbers in polar form if you please.

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Solution $2$

Let $z=r \operatorname{cis}(\theta)$. We want $z^2=i$, but $z^2=r^2 \operatorname{cis}(2\theta)$.

Now, $|i|=1$, $|z^2|=r^2$ and as we want $z^2=i$, we need $r^2=1$, i.e $r=1$.

Comparing angles now, we need $2\theta=90^\circ+360^\circ k$, that is, $\theta=45^\circ+180^\circ k$ or $\theta_1=45^\circ,\theta_2=225^\circ$ (note that if you add more $180^\circ$s, we get back $45,225,45,\cdots$.

So the two square roots of $i$ are $z_1=1\cdot \operatorname{cis}(45^\circ)$ and $z_2= 1\cdot \operatorname{cis}(225^\circ)$.

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I suggest you stop looking at your formulas and start thinking about the problem without them, as shown here, they're not needed, the only fact you 'needed to know' was that $i$'s angle is $90^\circ$.

Answer 4

In general, let $z=x+iy$ where $x$ and $y$ are real-valued numbers. We can express $z$ in Polar Coordinates as $z=\sqrt{x^2+y^2}e^{i\arctan(x,y)}$ where the Arctangent Function $\arctan(x,y)$ is given by

$$ \arctan(x,y)=\begin{cases} \arctan(y/x)&,x>0\\\\ \pi+\arctan(y/x)&,x<0,y>0\\\\ -\pi+\arctan(y/x)&,x<0,y<0\\\\ \pi/2&,x=0,y>0\\\\ -\pi/2&,x=0,y<0 \end{cases}$$

Then, the square root of $z$, $z^{1/2}$ is one of the two values

$$z^{1/2}=\pm (x^2+y^2)^{1/4}e^{i\arctan(x,y)/2} \tag 1$$

We can convert $z^{1/2}$ back to rectangular coordinates using Euler's Identity on $(1)$. Proceeding accordingly gives

$$\begin{align} z^{1/2}&=\pm (x^2+y^2)^{1/4}\left(\cos\left(\frac12 \arctan(x,y)\right)+i\sin\left(\frac12\arctan(x,y)\right)\right)\\\\ &=\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}\pm i\,\text{sgn}(y)\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\tag 2 \end{align}$$

where in arriving at $(2)$ we used the Half-Angle Formulae for the sine and cosine funcitons.

Applying $(2)$ to the case for which $x=0$ and $y=1$, we find that

$$\sqrt{i}=\pm\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)$$