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I'm working through "Calculus: Single Variable Part 1 - Functions" on Coursera platform and asked :

Find all possible solutions to the equation : $e^{ix} = i$

How to I begin with such a question ? I need to find the range of function of $e^{\sqrt-1x} = \sqrt-1$ , this means to find the range of values of x ?

I've researched these questions :

Finding all possible values

Finding all possible values of a Function

My conclusion is that while "Finding all possible solutions" is a common question the solutions to such type questions are unique.

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    Hint: Euler's Formula – 2017-02-18
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    What did you learn about the imaginary exponential, $e^{ix}$ ? – 2017-02-18

3 Answers 3

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$$e^{ix}= \cos x+ i \sin x $$

$$e^{ix}=i$$

So,$$\cos x+ i \sin x=i $$

$$\cos x=i(1-\sin x) $$

This is true only when both $\cos x=0$ and $\sin x=1$.

$x=\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\frac{-3\pi}{2},\frac{-7\pi}{2}, ...$

So, $x= (4n+1)\frac{\pi}{2}$ for integer $n$.

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    What is relationship between $\cos x=i(1-\sin x) $ and $x=\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\frac{-3\pi}{2},\frac{-7\pi}{2}, ...$ ? Is $\cos x=i(1-\sin x) $ "true only when both cosx=0 and sinx=1" ? – 2017-02-18
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    In the equation, a real LHS is equal to an unreal RHS. This can be only true when both sides eventually become zero. This is why both RHS and LHS need to be zero, which is possible only when $x= (4n+1)\frac{\pi}{2}$ or $x=\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},\frac{-3\pi}{2},\frac{-7\pi}{2}, ...$. – 2017-02-18
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    Now, look at curves of sin x and cos x simultaneously. The only time cos x = 0 and sin x = 1 is when x is of the form $(4n+1)\frac{\pi}{2}$ for some integer n. Hope this helps. – 2017-02-18
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    How do you arrive at computing product of $(4n+1)$ and $\frac{\pi}{2}$ ? I've looked at graphs of cos x = 0 and sin x = 0 , but I don't know how this relates to product of $(4n+1)$ and $\frac{\pi}{2}$ ? – 2017-02-18
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    $$\cos x=i(1-\sin x) $$ Notice that, for the above equation, one notable solution is $$x=\frac {\pi}{2}$$ Now, you know that adding $2\pi$ in any trigonometric function won't change the solution, ie, $\sin (2\pi+x)=\sin x$? Same goes for adding $4\pi$, $-2\pi$, and any multiple of $2\pi$. This means, other solutions can be $$x=\frac {\pi}{2} + 2\pi$$ $$x=\frac {\pi}{2} + 4\pi$$ $$x=\frac {\pi}{2} - 2\pi$$ We can generalize this to $$x=\frac {\pi}{2} + 2n\pi=\frac {\pi+4n \pi}{2}=(4n+1)\frac{\pi}{2}$$ for some integer $n$. – 2017-02-18
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    why pick $x=\frac{\pi}{2}$ as possible solution ? As $x=\frac{\pi}{2}$ relates to circle is it in order to "find all possible solutions"? – 2017-02-18
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    Naturally, we will look for solutions around zero. $$\sin x=1, \cos x=0$$ $$x = \frac{\pi}{2}$$ $x= \frac {\pi}{2}$ is the closest to zero. If we had taken $x= \frac {-3\pi}{2}$ or $x= \frac {5\pi}{2}$ as our base for other solutions, Then in generalization, $$x= (4n-3)\frac{\pi}{2}$$ and $$x= (4n+7)\frac{\pi}{2}$$ look a bit shabby as compared to $$x= (4n+1)\frac{\pi}{2}$$ Otherwise, they work just fine. We generally look for a simpler base that lies between $0$ and $2\pi$ (and sometimes $-\pi$ and $\pi$), which helps us easily evaluate other answers. – 2017-02-18
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    In fact, it makes no difference what we take as the particular solution, they all give literally the same general solution. – 2017-02-18
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We know that : $$e^{i\theta} = \cos \theta + i\sin \theta $$

Now, $$e^{i\frac {\pi}{2}}=? $$ Hope you can take it from here.

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Think geometrically. $e^{ix}$ represents the circle of radius 1 around the origin (every complex number can be written in polar form as $re^{i\theta}$ where $r$ is the modulus and $\theta$ is the angle with 0 in the positive real axis direction).

With this knowledge, what angle(s) $x$ will point you due north, where $i$ is?

Note: This is saying much the same thing as Euler's formula, since representing the exponential with sines and cosines is giving a polar representation.