This is a follow up to a previous question. I solved the equation $z^4 - 6z^2 + 25 = 0$ and I found four answer to be $z = \pm\sqrt{3 \pm 4i}$.

However someone in the comment said that the answer is going to be $2+i$, $2-i$, $-2+i$, $-2-i$. I cannot understand how we can find these roots from the answer that I found. How are we supposed to compute the square root of a complex number?

  • $\begingroup$ "I would like to know how we can approach this question using DeMoivre theorem?" This is not mentioned in the question, right? $\endgroup$ – Did Mar 29 '14 at 7:58
  • $\begingroup$ @Did You are right. In fact I was thinking that DeMoivre theorem is the only way to solve this until Ishfaaq show me an alternative way. $\endgroup$ – bman Mar 29 '14 at 9:39
  • $\begingroup$ Indeed, De Moivre could be the most ineffective approach here. $\endgroup$ – Did Mar 29 '14 at 9:41


Let $x + yi = \sqrt{a + bi}$. Then $(x+ yi)^2 = a + bi$. Then solve for $x$ and $y$ and you will generally have two sets of values for the square root $ \sqrt{a + bi}$


Say you want to compute $\sqrt{3 + 4i}$. Then assume the square root is $a + bi$. That is $a + bi = \sqrt{3 + 4i} \implies (a + bi)^2 = (a^2 - b^2) + 2abi = 3 + 4i$. Now solve the equations $ (a^2 - b^2) = 3$ and $2ab = 4$ to find $a$ and $b$.

  • $\begingroup$ Do you meant $x+yi=\sqrt{x^2+y^2}$ (modulus of a complex number)? $\endgroup$ – bman Mar 23 '14 at 3:32
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    $\begingroup$ He is assuming there is a complex number of the form $x+yi$ whose square root is $a+bi$ $\endgroup$ – ruler501 Mar 23 '14 at 3:35
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    $\begingroup$ It does not make sense to set $x+yi = \sqrt{x^2 +y^2}$ because this essentially says that the number is equal to it's magnitude. What you are doing is setting one complex number ($x+yi$) equal to the square root of another ($a+bi$) $\endgroup$ – Brad Mar 23 '14 at 3:36
  • $\begingroup$ Indeed you are right @Brad. I should have said $|x+yi| = \sqrt{x^2+y^2}$ $\endgroup$ – bman Mar 23 '14 at 3:39
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    $\begingroup$ We have to assume the square root too is a complex number so $(a + bi)$ is one such complex number where $i = \sqrt{-1}$. So $(a + bi)^2 = a^2 + (bi)^2 + 2abi$. Here, $(bi)^2 = b^2i^2 = b^2(\sqrt{-1})^2 = b^2(-1)$ $\endgroup$ – Ishfaaq Mar 23 '14 at 3:43

Let $$\sqrt{3+4i}=r(\cos\theta+i\sin\theta)$$ for some $r,\theta$. Squaring both sides, and using de Moivre, we get $$3+4i=r^2(\cos2\theta+i\sin2\theta)$$ so $$3=r^2\cos2\theta,\qquad4=r^2\sin2\theta$$ Squaring and adding gives $$25=3^2+4^2=r^4(\cos^22\theta+\sin^22\theta)=r^4$$ so $r=\sqrt5$. Dividing gives $$\tan2\theta=4/3$$ and $\theta=(1/2)\arctan(4/3)$. So, $$\sqrt{3+4i}=\sqrt5(\cos((1/2)\arctan(4/3))+i\sin((1/2)\arctan(4/3)))$$ This isn't exactly what you want, so let's have a look at that arctangent. Recall $$\tan2\theta={2\tan\theta\over1-\tan^2\theta}={2u\over1-u^2}$$ where I introduce the abbreviation $u=\tan\theta$. So, $${2u\over1-u^2}={4\over3}$$ This is $2u^2+3u-2=0$, $u=(-3\pm5)/4$. Let's take the plus sign, $u=1/2=\tan\theta$, so $\cos\theta=2/\sqrt5$, $\sin\theta=1/\sqrt5$. Now $$\sqrt{3+4i}=r(\cos\theta+i\sin\theta)=\sqrt5((2/\sqrt5)+i(1/\sqrt5))=2+i$$ With a little more care, you get the other three solutions.

It would seem de Moivre is not the best way to go, here.

  • $\begingroup$ You should know better than using $\sqrt{\ }$ applied to a complex number. $\endgroup$ – Did Mar 29 '14 at 9:18
  • $\begingroup$ @Did, what's wrong with $\sqrt z$? Yes, I understand that $2-i$ is on an equal footing with $2+i$; I'm happy to let OP work through the argument to see how to get both numbers. $\endgroup$ – Gerry Myerson Mar 29 '14 at 11:29
  • $\begingroup$ Well, I am a bit surprised to have to elaborate on this point but, to begin with, if $\sqrt{3+4i}=2+i$ and $\sqrt{3+4i}=-2-i$ (rather than your $2-i$, I guess) then $2+i=-2-i$ hence $1=0$. Thus one needs to specify that $\sqrt{3+4i}=2+i$ and $\sqrt{3+4i}\ne-2-i$. But this arbitrary convention leads to a function $\sqrt{\ }$ which cannot be continuous on the complex plane and such that $\sqrt{zw}$ is not always $\sqrt{z}\cdot\sqrt{w}$. All in all, the insistence on applying $\sqrt{\ }$ to complex numbers seems to have disastrous effects and very few, if any, positive ones. $\endgroup$ – Did Mar 29 '14 at 11:52
  • $\begingroup$ As an aside I fail to see what "you get the other three solutions" in your answer is referring to. $\endgroup$ – Did Mar 29 '14 at 11:54
  • $\begingroup$ @Did, first, of course you are right that I wrote $2-i$ but meant $-2-i$. But I maintain no harm is done provided one understands $\sqrt z$ as not a number but a (rather small) set of numbers. By "the other three solutions", I meant to the quartic in the body of the question. $\endgroup$ – Gerry Myerson Mar 29 '14 at 11:56

Suppose $(x+iy)^2 = a+ib$ with $a,b$ real and you want to find real values for $x$ and $y$.

Then $(x^2-y^2) + i(2xy) = a+ib$. Since $a,b,x,y$ are real, this is equivalent to $x^2-y^2 = a$ and $2xy = b$

$(x^2+y^2)^2 = (x^2-y^2)^2 + 4x^2y^2 = a^2+b^2$, so $x^2+y^2 = \sqrt {a^2+b^2}$. You always pick the positive root because $x^2+y^2$ is positive.
Then, $x^2 = \frac {\sqrt {a^2+b^2}+a}2$, so that $x = \pm \sqrt \frac {\sqrt {a^2+b^2}+a}2$, and finally $y = \frac b{2x}$ if $x \neq 0$.
If $x = 0$ you have to use $y = \pm \sqrt \frac {\sqrt {a^2+b^2}-a}2$ instead.

Here, for $a=3, b=4$, $a^2+b^2 = 9+16 = 25 = 5^2$ so that $x = \pm \sqrt {\frac{5+3}2} = \pm 2$, and $y = \frac 4 {2x} = \pm 1$ (with the same sign as $x$). This gives the two square roots of $3+4i$ .

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    $\begingroup$ But the request was for an approach using de Moivre. $\endgroup$ – Gerry Myerson Mar 29 '14 at 8:37

How are we supposed to compute the square root of a complex number?

If $z$ is a nonpositive real number, the square roots are $\pm\sqrt{-z}\cdot\mathrm i$. Otherwise, the square roots of $z$ are

$$ \pm\sqrt{r}\cdot\frac{z+r}{|z+r|},\qquad\text{with}\ r=|z|. $$

The formula is legit because $z+r\ne0$. The whole procedure requires to use the usual operations $+$, $-$, $\times$, $\div$, and to compute thrice the square root of some positive real number... but nothing else.

Example: If $z=3+4\mathrm i$ then $r^2=3^2+4^2$ hence $r=5$, $z+r=8+4\mathrm i$, that is, $z+r=4(2+\mathrm i)$, $|z+r|=4|2+\mathrm i|=4\sqrt5$ and the square roots are $$ \pm\sqrt5\cdot\frac{4(2+\mathrm i)}{4\sqrt5}=\pm(2+\mathrm i). $$ For a somewhat less made up example, consider $z=3+2\mathrm i$, then the identity above shows that the square roots of $z$ are $$ \pm\frac{3+\sqrt{13}+2\mathrm i}{\sqrt2\cdot\sqrt{3+\sqrt{13}}}. $$

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    $\begingroup$ Have you used de Moivre here, as requested? $\endgroup$ – Gerry Myerson Mar 29 '14 at 8:38
  • $\begingroup$ @GerryMyerson You cannot say? I am surprised... (See comment to main answer.) $\endgroup$ – Did Mar 29 '14 at 9:17
  • $\begingroup$ I don't know what you mean by "main answer". There is one answer with many upvotes, but it also has many comments, none from you. Anyway, OP predicated the bounty on an answer using de Moivre. I agree that de Moivre is less than optimal here, but they say the customer is always right. $\endgroup$ – Gerry Myerson Mar 29 '14 at 11:34
  • $\begingroup$ Sorry: *main question. $\endgroup$ – Did Mar 29 '14 at 11:55

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