# Square root of i ? (and another math kweshun) **URL:** \ **Category:** Factual Questions\ **Created:** [February 26, 2004, 8:23am UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262 "2004-02-26T08:23:24Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Ackafool\_Upnhere](https://avatars.discourse-cdn.com/v4/letter/a/22d042/32.png) [@Ackafool\_Upnhere](https://boards.straightdope.com/u/Ackafool_Upnhere)\ **Post date:** [February 26, 2004, 9:11pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/21 "2004-02-26T21:11:26Z")
> [@Chronos](#): > > There are a lot of things you’re not allowed to do, in the real numbers. You can’t take a log of a negative number, you can’t take the square roots of negative numbers, you can’t take the inverse sine of a number greater than 1 or less than -1, etc. But _all_ of these problems disappear when you’re working with complex numbers. Not only can you take a log of a negative number, you can even take the log of an imaginary number, or any complex number (except zero). You can take any complex number to any power, or to any root. You can do any trig function or inverse trig function you like. And you never need to introduce any system of numbers larger than the complex numbers. That’s the great beauty of the complex numbers. **Chronos** , you got me all excited about all these things you can do in the complex world, which are not possible in the real world. Just curious: What does an angle with a sin \>1 look like on my trusty trig circle? Can you take the log of a negative real number if you write it as the product of a positive number and i^2? what is the log of i? Gosh, did I tell you you got me all excited? 😉 ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [February 26, 2004, 9:14pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/22 "2004-02-26T21:14:57Z")
> [@Ackafool Upnhere](#): > > So, going back to my trig circle, the process of squaring a complex number just a matter of squaring its module and doubling its argument, right? Does this extend to any powers of a number? And conversely, extracting the nth root of a complex number will yield n solutions? That’s correct. In general, the n roots of a number will be the vertices of a regular n-gon centered at the origin. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [February 26, 2004, 9:50pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/23 "2004-02-26T21:50:45Z")
One other thing I’ve always liked is that **i[sup]i[/sup]** is real. (It’s equal to exp(-pi/2), or 0.2089796.) ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [February 26, 2004, 10:21pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/24 "2004-02-26T22:21:22Z")
I’ve a soft spot for Euler’s “mathematical poem” myself: _e[sup]i p [/sup] + 1 = 0_. Five fundamental constants, addition, subtraction, exponentiation and equality all rolled into one tiny equation. Shame he couldn’t fit division in there as well! ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [February 26, 2004, 10:31pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/25 "2004-02-26T22:31:33Z")
Eh… for “subtraction” read “multiplication”, of course. Btw, concerning the question about _nth_ roots, there are _n_ such roots of **any** number, real, imaginary or complex. For instance, the humble number 1 has three cube roots, two of which are complex. You should be getting the idea of how to evaluate these by now. They have some interesting properties when added, subtracted and multiplied 🙂 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [February 26, 2004, 10:52pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/26 "2004-02-26T22:52:59Z")
> [@Ackafool Upnhere](#): > > Just curious: What does an angle with a sin \>1 look like on my trusty trig circle? > Can you take the log of a negative real number if you write it as the product of a positive number and i^2? what is the log of i? Gosh, did I tell you you got me all excited? 😉 In higher math, the sine as the ratio of sides of a triangle is not used. Rather, sin(x) = x + x[sup]3[/sup]/3! + x[sup]5[/sup]/5! + … A number whose sine is greater than 1 is simply a number whose series gets big a little faster than any real number’s. As far as logarithms go… log(c) is defined for any complex number c, including the negative reals. The problem has to do with the polar representations. (r, t) and (r, t + 2[symbol]p[/symbol]) represent the same point, but log((r, t)) and log((r, t + 2[symbol]p[/symbol])) are not equal. Things get ugly because of this. ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [February 26, 2004, 11:01pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/27 "2004-02-26T23:01:29Z")
The poor OP was just beginning to learn about imaginary numbers, folks! It’s like she asked for a cup of coffee and received instruction on how to make it starting with the bean! Forgetting polar coordinate and Euler (some pronounce “Oiler”) So, “Oiler vay”, Batman! Give the poor OP a break! To me, imaginary numbers simply indicate there are cracks in the theory behind the very numbering system we use. (Of course, no system is perfect.) I say this because, as so many posts above are trying to show, those so-called imaginary numbers have very REAL application, in the engineering underworld! - Jinx 😃 ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [February 26, 2004, 11:14pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/28 "2004-02-26T23:14:12Z")
If you want a number with a weird sine, use this formula: sin(x + **i** y) = sin(x) cosh(y) + **i** cos(x) sinh(y) So, the unwieldy pi/2 - **i** ln(sqrt(3) + 2) = 1.570796 - 1.316958 **i** has a sine of 2. > [@Jinx](#): > > The poor OP was just beginning to learn about imaginary numbers, folks! It’s like she asked for a cup of coffee and received instruction on how to make it starting with the bean! Forgetting polar coordinate and Euler (some pronounce “Oiler”) So, “Oiler vay”, Batman! Give the poor OP a break! Hey! The reason I went into the complex plane is that OP mentioned it! “picturing the trigonometric circle where i is the segment sticking up in the air and -1 (i^2) is the segment going ‘straight and to the left’…where would i^1/2 be ?” And it’s a good thing, too. Complex roots are much simpler in terms of their polar form. ---
**Author:** ![Ackafool\_Upnhere](https://avatars.discourse-cdn.com/v4/letter/a/22d042/32.png) [@Ackafool\_Upnhere](https://boards.straightdope.com/u/Ackafool_Upnhere)\ **Post date:** [February 26, 2004, 11:28pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/29 "2004-02-26T23:28:54Z")
> [@Jinx](#): > > The poor OP was just beginning to learn about imaginary numbers, folks! It’s like she asked for a cup of coffee and received instruction on how to make it starting with the bean! Forgetting polar coordinate and Euler (some pronounce “Oiler”) So, “Oiler vay”, Batman! Give the poor OP a break! > > To me, imaginary numbers simply indicate there are cracks in the theory behind the very numbering system we use. (Of course, no system is perfect.) I say this because, as so many posts above are trying to show, those so-called imaginary numbers have very REAL application, in the engineering underworld! - Jinx 😃 Well yeah, it’s scary- but it’s a GOOD kind of scary. It’s almost like finding religion when I look into such things… Thanks again everybody for widening the ol’ horizon. I have to spend a few eons now digesting all that has been said here, especially the examples. Feel free to bring more on! ---
**Author:** ![Ackafool\_Upnhere](https://avatars.discourse-cdn.com/v4/letter/a/22d042/32.png) [@Ackafool\_Upnhere](https://boards.straightdope.com/u/Ackafool_Upnhere)\ **Post date:** [February 26, 2004, 11:48pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/30 "2004-02-26T23:48:43Z")
> [@Malacandra](#): > > Eh… for “subtraction” read “multiplication”, of course. > > Btw, concerning the question about _nth_ roots, there are _n_ such roots of **any** number, real, imaginary or complex. For instance, the humble number 1 has three cube roots, two of which are complex. You should be getting the idea of how to evaluate these by now. They have some interesting properties when added, subtracted and multiplied 🙂 OK **Malacandra** , you are the cause of my current headache- I can visualize the 3 cubic roots of 1 on my trig circle, they have to be 1, e^i_π/3 and e^i_2π/3. Now, I can’t seem to figure out how to solve this from equations. Help, someone! Starting from: (x+iy)³=1 x³+3x²iy+3x(iy)²+(iy)³=1 x³+3ix²y-3xy²-iY³=1 which implies x³-3xy²=1 and 3x²y-y³=0 I’m stuck. ☹ ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [February 26, 2004, 11:58pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/31 "2004-02-26T23:58:01Z")
> [@Ackafool Upnhere](#): > > OK **Malacandra** , you are the cause of my current headache- I can visualize the 3 cubic roots of 1 on my trig circle, they have to be 1, e^i_π/3 and e^i_2π/3. Not quite. They are exp(2[symbol]pi[/symbol]/3), exp(4[symbol]pi[/symbol]/3), and exp(6[symbol]pi[/symbol]/3), which is equal to 1. In general, the nth roots of one are given by exp(2k[symbol]pi[/symbol]/3), with 0 \< k \< n - 1. > [@](#): > > Now, I can’t seem to figure out how to solve this from equations. Help, someone! > > Starting from: > > (x+iy)³=1 > > x³+3x²iy+3x(iy)²+(iy)³=1 > > x³+3ix²y-3xy²-iY³=1 which implies > > x³-3xy²=1 and 3x²y-y³=0 > I’m stuck. ☹ Once you know the formula above, there’s no reason to try to solve the equations–they’re nasty. ---
**Author:** ![Ackafool\_Upnhere](https://avatars.discourse-cdn.com/v4/letter/a/22d042/32.png) [@Ackafool\_Upnhere](https://boards.straightdope.com/u/Ackafool_Upnhere)\ **Post date:** [February 27, 2004, 12:02am UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/32 "2004-02-27T00:02:15Z")
Yes! I gotcha, Ultrafilter. In a momentary lapse of reason, I had the 180 degree angle on my circle equal π/2, not π (which halved all my roots). Thanks! ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [February 27, 2004, 12:03am UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/33 "2004-02-27T00:03:30Z")
It’ll also help to know that: exp( **i** x) = cos(x) + **i** sin(x) In general, if the polar coordinates of any complex number are given by (r, theta), then the complex number can be represented as r exp( **i** theta). ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [February 27, 2004, 11:38am UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/34 "2004-02-27T11:38:28Z")
> [@Ackafool Upnhere](#): > > Now, I can’t seem to figure out how to solve this from equations. Help, someone! > > Starting from: > > (x+iy)³=1 > > x³+3x²iy+3x(iy)²+(iy)³=1 > > x³+3ix²y-3xy²-iY³=1 which implies > > x³-3xy²=1 and 3x²y-y³=0 > I’m stuck. Allow me to unstick you, then. Factorise your equations: x(x[sup]2[/sup] - 3y[sup]2[/sup]) = 1 y(3x[sup]2[/sup] - y[sup]2[/sup]) = 0 The second equation has two solutions: y=0, or y[sup]2[/sup]=3x[sup]2[/sup]. (Actually the second of these implies two solutions, but we’ll come back to that). Going back to the first equation and substituting 3x[sup]2[/sup] for y[sup]2[/sup], we get: -8x[sup]3[/sup] = 1 and so x = -1/2 Given that y[sup]2[/sup]=3x[sup]2[/sup], there are two solutions: y=sqrt(3)x and y = -sqrt(3)x, i.e. sqrt(3)/2 and -sqrt(3)/2 So this gives us the three cube roots of 1: 1 -1/2 + (sqrt(3)/2)i -1/2 - (sqrt(3)/2)i The complex roots are known as w and w[sup]2[/sup], and as I said, you can have fun adding and multiplying the roots together 🙂 From here on in the _n_th roots get hairy, and you’re better off with the “polar coordinates” approach. Btw, this post was a sod to code!! ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [February 27, 2004, 3:20pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/35 "2004-02-27T15:20:34Z")
> [@Malacandra](#): > > I’ve a soft spot for Euler’s “mathematical poem” myself: _e[sup]i p [/sup] + 1 = 0_. Five fundamental constants, addition, subtraction, exponentiation and equality all rolled into one tiny equation. Shame he couldn’t fit division in there as well! I know you meant multiplication instead of subtraction, but it’s relatively easy to divide zero by the square root of one minus v-squared over c-squared. 🙂 ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [February 27, 2004, 7:13pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/36 "2004-02-27T19:13:32Z")
One point that should be made is that all this talk about logs of negative numbers and arc sines of numbers larger than 1 have to be taken with a grain of salt. Just as i (like any non-zero number) has two square roots, any number has infinitely many logarithms! The fact that e^2_pi_i = 1 implies that you can add an arbutrary integer multiple of 2_pi_i to any logarithm. Sure log (-1) = pi_i, but it also equals -pi_i and 3_pi_i and -7_pi_i and … BTW, not only does every complex number have n nth roots, any polynomial equation of degree n with complex number coefficients has n solutions (although sometimes you get repeated roots, such as x^2 - 2x + 1 = 0 has 1 as a double root). ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [February 28, 2004, 7:07am UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/37 "2004-02-28T07:07:23Z")
> [@Hari Seldon](#): > > One point that should be made is that all this talk about logs of negative numbers and arc sines of numbers larger than 1 have to be taken with a grain of salt. Just as i (like any non-zero number) has two square roots, any number has infinitely many logarithms! The fact that e^2_pi_i = 1 implies that you can add an arbutrary integer multiple of 2_pi_i to any logarithm. Sure log (-1) = pi_i, but it also equals -pi_i and 3_pi_i and -7_pi_i and … I don’t think it’s that big of a problem. It’s not unlike the arcsine function on the real numbers. Sure, there are an infinite number of solutions to sin(x) = 0.4, but we still define arcsin(0.4) as the “primary” value. ---
**Author:** ![DarrenS](https://avatars.discourse-cdn.com/v4/letter/d/977dab/32.png) [@DarrenS](https://boards.straightdope.com/u/DarrenS)\ **Post date:** [February 28, 2004, 4:27pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/38 "2004-02-28T16:27:36Z")
> [@Achernar](#): > > One other thing I’ve always liked is that **i[sup]i[/sup]** is real. (It’s equal to exp(-pi/2), or 0.2089796.) Now _that_ is freaky - I never knew that. How can one rationalize (pun unavoidable) that? Take the unitary imginary number, raise it to the power of itself, and you get something that is not only real but some weird, probably transcendental thing. I can’t get my intuition around that. I know you can probably derive it logically from one of the identities mentioned in this thread, but I like to have an intuitive feel for things. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [February 28, 2004, 5:07pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/39 "2004-02-28T17:07:22Z")
> [@Achernar](#): > > I don’t think it’s that big of a problem. It’s not unlike the arcsine function on the real numbers. Sure, there are an infinite number of solutions to sin(x) = 0.4, but we still define arcsin(0.4) as the “primary” value. Even if you do principal values, you still have to worry about branch cuts. > [@DarrenS](#): > > Now that is freaky - I never knew that. How can one rationalize (pun unavoidable) that? Take the unitary imginary number, raise it to the power of itself, and you get something that is not only real but some weird, probably transcendental thing. I can’t get my intuition around that. I know you can probably derive it logically from one of the identities mentioned in this thread, but I like to have an intuitive feel for things. i[sup]i[/sup] = exp(i \* ln(i)) = exp(i/2 \* ln(-1)) = exp(i/2 \* i[symbol]p[/symbol]) = exp(-[symbol]p[/symbol]/2) It’s hard to get an intuitive feel for complex exponentiation. Consider that I could’ve chosen any other branch of the logarithm and gotten a different answer. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [February 28, 2004, 7:28pm UTC](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262/40 "2004-02-28T19:28:19Z")
> [@ultrafilter](#): > > Even if you do principal values, you still have to worry about branch cuts. > > i[sup]i[/sup] = exp(i \* ln(i)) = exp(i/2 \* ln(-1)) = exp(i/2 \* i[symbol]p[/symbol]) = exp(-[symbol]p[/symbol]/2) > > It’s hard to get an intuitive feel for complex exponentiation. Consider that I could’ve chosen any other branch of the logarithm and gotten a different answer. You have to be careful there. I know that is your point, but the equality doesn’t necessarily hold. It’s the same reason that we don’t say -2 = Sqrt(4), because we know that 2 = Sqrt(4) and if that first were true equality, that would imply -2 = 2, since they’d both be equal to the same “thing”. [Previous page](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262.md?page=1) [Next page](https://boards.straightdope.com/t/square-root-of-i-and-another-math-kweshun/231262.md?page=3)