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Very simple question with an answer that I cannot understand:

I have $\sqrt{8i}$, which, I suppose, is the same as $\sqrt{\sqrt{-64}}$. How come that $2+2i$ is the same as $\sqrt{8i}$?

My understanding is that $\sqrt{8i}$ is the same as: (a) $\sqrt{2^3i}$ OR (b) $2\sqrt{2i}$

I'm pretty sure (a) is correct and (b) might also be correct, but how can you get from there to $2+2i$?

Thanks in advance

mrf
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complexguest
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    I think you need to know what is $\sqrt{i}$ first. It seems to me that you can find it in the site cause it was certainly asked before here. – Mikasa Jul 28 '13 at 11:06
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    Every complex number can be put in the form $a+bi$. Set $(a+bi)^2=8i$ and try solving for $a$ and $b$ (each which must be real). In general polar form of numbers is useful. – anon Jul 28 '13 at 11:08
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    Multiplying complex numbers means multiplying absolute values and adding angles, i.e. rotating. The currently "accepted" answer to this question ignores the simple geometry. See my answer below. – Michael Hardy Jul 28 '13 at 14:17

8 Answers8

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one way to see it is there exists some $z$ such that $z^2=8i$, now sub $z=a+ib$ to get $a^2-b^2+2iab=8i$, equate real and the imaginary parts, solve the simultaneous equation. if you know about the exponential form of complex numbers, this problem can be solved even quicker but I suspect you haven't learnt that yet. By the way be careful in the above statement when you say $\sqrt{\sqrt{}}$, personally its not evidently clear your allowing both the $\sqrt{}$ values to be accounted as it seems your accounting for the positive only, remember the $\sqrt{}$ of negatives don't exist in the reals but does in complex numbers.

Samrat Mukhopadhyay
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WhizKid
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  • Assuming that 8i = 0+8i, then I will have: a^2-b^2 = 0 and 2iab = 8i; I can solve the first and conclude that a^2 = b^2; then I go to the second, and if I solve for a, I get a = 4/b; if I solve for b I get b = 4/a. At this point I substitute a=4/b to a^2 = b^2 and I get (4/b)^2 = b^2 which is 16/b^2 = b^2. From there I multiply both sides for b^2 and I get 16 = b^4, which means that b +/- 2 (and the same applies for a, I suppose). At this point I undestand that 4-4+222i = 8i, but I still don't get how to arrive to sqrt(8i) = 2+2i :( – complexguest Jul 28 '13 at 11:38
  • 8i=0+8i, dont assume that, its the complex representation. Now yes you have a^2=b^2 which means a=modb, and yes with the second equation you have a=4/b, sub that into the first equation and multiply by b^2 to get b^4=16, this implies b=plus or minus 2, then for each value of b, check the value of a, you will get + and - (2+2i). Remember the idea of doing this is that a,b are real, so essentially we are solving this problem by going to equations with real solutions for a and b, as a and b are defined to be real. – WhizKid Jul 28 '13 at 12:02
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HINT: $(1+i)^2=1+2i+i^2=2i\qquad\qquad$

Brian M. Scott
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let $$x+iy=\sqrt{8i}$$ $$(x+iy)^2={8i}$$ $$x^2-y^2+2xyi=0+8i$$ compare real and imaginary parts of both sides: $x^2-y^2=0\;; 2xy=8$ $$x^2-y^2=0$$ square both sides $$\implies (x^2-y^2)^2 =0\implies (x^2+y^2)^2-(2xy)^2=0$$ $$\implies (x^2+y^2)^2=64 \implies (x^2+y^2)=\pm 8 $$ since sum of square of two real no. can't be negative so $$x^2+y^2=8\text{ and }x^2-y^2=0$$ solving these two eqn will give $x=y=\pm 2 $

so $$\sqrt {8i}=\pm 2(1+i)\implies 2+2i,-2-2i$$

Michael Hardy
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iostream007
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You can go this way

$$ \sqrt{8} \sqrt{i}= \sqrt{8} e^{\frac{1}{2}\ln i}= \sqrt{8}\,e^{\frac{1}{2}\left(\ln|i|+i\left(\frac{\pi}{2}+2k\pi\right)\right)}= \sqrt{8}\,e^{\left(i(\frac{\pi}{4})+k\pi\right)},\quad k=0,1.$$

Note that, if you take $k=2,3,\dots$, then you go back to the same roots.

Mhenni Benghorbal
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    Using Log $i;$ for such basic questions is pretty risky or even plain wrong: the complex logarithmic function is infinitely valued and unless one definite branch is chosen for it there is no way to tell what Log $i;$ is, though I'm guessing you meant the usual branch deleting all the non-positive real axis. Anyway, it is higly doubtable someone asking this question can grasp all this. – DonAntonio Jul 28 '13 at 11:39
  • @DonAntonio: Thanks for the comment. – Mhenni Benghorbal Jul 28 '13 at 11:56
  • @DonAntonio: See my answer to this problem which is related to what you said. – Mhenni Benghorbal Jul 28 '13 at 12:04
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    I was reading answers and thinking "What's happening? Why there isn't anyone with simple and clear trigonometric formula?". And here you are :) – tkroman Jul 28 '13 at 15:26
  • @cdshines: Thank you for the comment. I really appreciate it. – Mhenni Benghorbal Jul 28 '13 at 15:28
  • Your bracketing is I think wrong in your final part of the equality (the i should be multiplying everything I think? and in the previous part its doing strange things with big and little brackets... – Chris Jul 28 '13 at 21:54
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    @Chris: Fixed. Thanks for the comment. – Mhenni Benghorbal Jul 29 '13 at 01:57
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The absolute value of $8i$ is $8$, so the absolute value of its square root is $\sqrt{8} = 2\sqrt{2}$.

The angle from the real axis to the imaginary axis (where $8i$ is located) is $90^\circ$, so the angle from the real axis to the ray on which $\sqrt{8i}$ is located is half of that: $45^\circ$. Consequently it's a number of the form $a+ai$, where $a>0$. The absolute value of $a+ai$ is $\sqrt{a^2+a^2} = a\sqrt{2}$.

So we have $a\sqrt{2} = 2\sqrt{2}$, so $a=2$.

Michael Hardy
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Hint :

$$ \sqrt i = \frac1{\sqrt2}(1+i) $$

Mahdi Khosravi
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$(a+bi)^2=(a+bi)(a+bi)=a^2-b^2+2abi\implies \sqrt{a^2-b^2+2abi}=a+bi$
and for conjugate $(a-bi)^2=(a-bi)(a-bi)=a^2+b^2-2abi\implies \sqrt{a^2+b^2-2abi}=a-bi$
Now if we find $a$ and $b$ satisfy appropriate equotation:
For $(a+bi)^2$:
$a^2-b^2+2abi=8i$. This is valid for any $a=b$ where $2ab=8$, or
$2a^2=8 \implies a=\pm\sqrt{4}=\pm{2} = b$. Here exists two numbers:
First $$\sqrt{8i}=2+2i$$ and second $$\sqrt{8i}=-2-2i$$
For $(a-bi)^2$:
$a^2+b^2-2abi=8i$. Because $a^2 \ge 0$ and $b^2 \ge 0$, there does not exists solution.

Stefan Babos
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  • Your conjugate should be -b^2, not +b^2. The reason being that its -bi-bi = bibi = -b^2. The correct interpretation also of a^2-b^2 is that a = b or a=-b. In the first case ab=4 means a=-b is not possible. In your conjugate example ab=-4 and thus a=-b. It gives the same answer whether you use a+bi or a-bi, its just the sign of b is changed. – Chris Jul 28 '13 at 21:48
  • This answer also avoids the geometry. – Michael Hardy Jul 29 '13 at 05:14
  • Upvoters: why the upvote? – Did Jul 31 '13 at 15:54
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First note that $(2+2i)^2=4-4+8i=8i$. Now to be exact $$\sqrt{8i}=2\sqrt{2}e^{i(2k+1)\pi/4}\tag{*}$$ where $k\in \{0,3\}$. So actually $\sqrt{8i}$ is multi-valued just as $\sqrt{8}=\pm 2\sqrt{2}$ is multi-valued.. The identity $(*)$ comes from a famous formula proved by Euler. You can learn more about it here.

Samrat Mukhopadhyay
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    You can't have $4$ solutions... All solutions are roots of $X^2-8i$ which is of degree $2$ so you can have at most $2$ solutions... In fact, you can factor it as $(X+\alpha)(X-\alpha)$ for any $\alpha$ so that $\alpha^2 = 8i$. – xavierm02 Jul 28 '13 at 11:16