# How to solve $e^{ix} = i$

I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help.

Thanks.

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the important thing to notice here is that there is more than one $x$ which works. – user27182 May 28 '13 at 22:24
Just curious: what online course? If it's free, there might be others (such as myself) who may be interested... :) – apnorton May 28 '13 at 22:52
@anorton It is called "Calculus: Single Variable" and it can be found on coursera.org – Jeel Shah May 28 '13 at 22:54

So, you say that you're okay with Euler's formula: $$e^{ix}=\cos(x)+i\sin(x),$$ and it should be clear that (for real numbers $a,b,c,d$), we have $$a+bi=c+di\quad\iff\quad a=c\;\text{ and }\;b=d.$$ So, which values of $x$ will satisfy $$\cos(x)+\sin(x)i=0+1i\quad ?$$

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To tack onto this answer.. there will be many such values of $x$ that solve the above equality. – Cameron Williams May 28 '13 at 22:27
@ZevChonoles how do you know that $\cos(x)+\sin(x)i=0+1i\quad$ Where did you find $0+1i$? Why are we assuming that $\cos(x)$ is zero and $\sin(x)$ is one? – Jeel Shah May 28 '13 at 22:28
@gekkostate: that is the point of the two equivalent equations. We know $\cos x$ and $\sin x$ are real for real $x$ so you can identify real and imaginary parts. – Ross Millikan May 28 '13 at 22:40
@gekkostate: I am just restating the problem; it is not something I somehow "know". You're given the equation $$e^{ix}=i$$ and told to find the values of $x$ for which this is true. But $$e^{ix}=\cos(x)+i\sin(x)$$ and $$i=0+1i$$ so you are, equivalently, finding the values of $x$ for which $$\cos(x)+i\sin(x)=0+1i.$$ – Zev Chonoles May 28 '13 at 22:45
@gekkostate if I have no reals and one $i$, then I just have $i$. $0+1i=i$. – Robert Mastragostino May 28 '13 at 23:03

$$e^{ix} = i$$ Euler's formula: $$e^{ix}=\cos(x)+i\sin(x),$$ so: $$\cos x+i\sin x = 0+1\cdot i$$ compare real and imaginary parts $$\sin x =1$$ and $$\cos x =0$$ $$x=\dfrac{(4n+1)\pi}{2}\;\;,n \in \mathbb W$$ (W stands for set of whole number $W=${$0,1,2,3,.......,n$}).

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Shouldn't it bee $4n+1$ in the numerator? With $2n+1$ I get negative values – Alex May 28 '13 at 22:29
yeah it should thanks @Alex – iostream007 May 28 '13 at 22:36
@iostream007 Can you please explain the first two lines? How does $e^{ix} = i$ translate to $\cos x+i\sin x = 0+1\cdot i$? – Jeel Shah May 28 '13 at 22:39
@gekkostate Euler's formula: – iostream007 May 29 '13 at 4:55
$e^{ix}=\cos x+\sin x$ comes from euler's formula and $i$ has written as $0+1\cdot i$ – iostream007 Jun 9 '13 at 16:17

Well, the geometric meaning of $e^{ix}$ is the weapon here to use.

This is nothing but the point on the complex plane which has length $1$ and angle $x$ measured from the right half of the real axis, in radian. So that $e^{i\pi}=-1$, for example.

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-1: This is the best representation, but from my experience a student using this method will only give one answer. – Were_cat May 29 '13 at 9:32
But the student will see, that's more important for long term.. :) – Berci May 29 '13 at 11:09

A completely different approach:

$$e^{i\pi} = -1 \implies \sqrt {e^{i\pi}} = \sqrt {-1} \implies e^{\frac{i\pi}{2}} = i$$

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Can you explain why you chose the 'positive' value of the square root? – Calvin Lin May 29 '13 at 0:52

The key here is to think in terms of radial co-ordinates rather than cartesian. Whenever you see a complex exponential it's best to think of the geometric interpretation first, if possible.

The radial representation of a complex number $z = a + bi$ is $Re^{i\theta}$
where $R = |z|$ and $\theta=arctan(b/a)$.

So for the number i we ask ourselves what is R, and what is $\theta?$
We look where the point i is on the complex plane - it's on the y axis 1 unit away from the origin. So R is 1 and $\theta$ is $\frac{\pi}{2}$. (Note that you can add any multiple of $2\pi$ to $\theta$ and things stay the same.)

So plug these R, $\theta$ values into the radial form of z to get the radial representation for i which is $1.e^{i\frac{\pi}{2}} = e^{i\frac{\pi}{2}}$
So $e^{ix} = i = e^{i\frac{\pi}{2}}$ and a naive answer is $x = \frac{\pi}{2}$. But remember we could add any multiple of $2{\pi}$ to our angle for i so we need to add $2n{\pi}$ for the general answer.

Hence $x = \frac{\pi}{2}+2n\pi \quad n\in \mathbb{Z}$

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$$e^{ix} = i \implies ix =\ln(i) \implies ix= \ln(|i|)-i \left(\frac{\pi}{2}+2k\pi\right)$$

$$\implies x = \frac{\pi}{2}+2k\pi,\quad k\in \mathbb{Z}.$$

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Hint:

$$e^{xi}=i=e^{\frac\pi2i+2k\pi i}\implies x=\frac\pi2+2k\pi\;,\;\;k\in\Bbb Z$$

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