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I am in high school, and we started learning De Moivre's formula. I had some problems with my homework concerning rooting of z. So far, this is what I know about the formula:

$\sqrt[n]{z}= \sqrt[n]{r}\left (\cos \left(\dfrac{\phi+2k\pi}{n} \right)+i\sin \left(\dfrac{\phi+2k\pi}{n} \right) \right)$

where $z = a + bi$ and $r = \sqrt{a^2 + b^2}$

Now here's the problem. What is $\phi$ equal to? My professor told us that $\phi = \arctan \left(\dfrac ba \right)$ but this seems incorrect to me. In my homework I had a simple equation of $x^2 + 1 = 0$ where $x = \sqrt{-1}$ and if you try solving $\phi$ with the above formula, you will get it to be 0, whereas the correct answer is $\phi = \pi$

So is $\phi = \arctan \left(\dfrac ba \right)$ a wrong way to find $\phi$, or am I mistaking somewhere?

Nεo Pλατo
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David Simka
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    As lab bhattacharjee has said, there are two values $\phi$ can be. Think about the graph of the tangent function, the part from $0$ to $\pi$ looks just like the part from $\pi$ to $2\pi$. You have to figure out which one it is by using the signs of $a$ and $b$. (When you divide them, some data is lost, you can't tell the difference between $+/+$ and $-/-$). You may also want to look at atan2. – Henry Swanson Sep 10 '13 at 04:31

3 Answers3

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$\phi$ must be an angle such that $a = r\cos\phi$ and $b = r\sin\phi.$

The value of the arc tangent function is strictly between $-\dfrac\pi2$ and $\dfrac\pi2$ for every possible input. That is, it can only produce an angle in the fourth quadrant, in the first quadrant, or on the positive real number line. To put it another way, if $\theta$ is the result of $\arctan\left(\dfrac ba\right),$ then $\cos\theta > 0.$

Now, this causes a difficulty for you because $r \geq 0$ and (in the problem you are looking at) $a = -1.$ From $a = r\cos\phi$ we conclude $\cos\phi < 0.$ Therefore $\phi$ cannot be the result of $\arctan\left(\dfrac ba\right).$

The solution to your difficulty is that if $\theta = \arctan\left(\dfrac ba\right),$ then $$\tan(\theta + \pi) = \tan(\theta - \pi) = \tan(\theta) = \dfrac ba.$$

But if $\cos(\theta) > 0,$ then $\cos(\theta + \pi) = \cos(\theta - \pi) < 0.$ So if you need an angle with a negative cosine (which you do in this exercise), try $\arctan\left(\dfrac ba\right) + \pi$ or $\arctan\left(\dfrac ba\right) - \pi.$

In this case $\arctan\left(\dfrac ba\right) + \pi = \pi,$ and $\phi = \pi$ works very well.

David K
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If you write a complex number $a+bi$ in polar form $r(\cos\phi+i\sin\phi),$ then you must have $a=r\cos\phi$ and $b=r\sin\phi.$ Dividing these equations gives $$\frac ba=\tan\phi,$$ where $r\ne 0.$ So yes, you have that $$\phi=\arctan\left(\frac ba\right).$$

Allawonder
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    Assuming that $r > 0$ (without which the method doesn't work in general), this answer is only correct when $a > 0.$ And in this question, $a = -1.$ – David K May 26 '24 at 06:31
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In fact $,\phi=\arctan \frac {-1}0$ as $b<0,a=0$ $\phi $ will lie in the Third Quadrant

For more example, let $z=-1-i,$

$\arctan \frac{-1}{-1}\ne \arctan \frac11$ as $b<0,a<0,\phi $ will lie in the Third Quadrant

The details can be found here

There is a range of values $n$ that can assume, that is $0,1,2\cdots, n-1$ (explained here) so as to give $n$ roots of $x^n=r$

Here $\displaystyle x^2=-1=\cos(2n\pi+\pi)+i\sin(2n\pi+\pi)$

$$\implies x=\cos\frac{(2n+1)\pi}2+i\sin\frac{(2n+1)\pi}2=i\sin\frac{(2n+1)\pi}2$$ where $n=0,1$

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lab bhattacharjee
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