# Complex operator $i$ and Exponents

I am trying to understand the complex numbers and exponents.

I came across this question. I wonder how to explain the difference between $${2\cdot i} \text{ and } 2^i$$ as $$i=\sqrt{-1}$$

edit:

Rather than explaining the meaning of above two numbers in yet another equally difficult mathematical form.. i am more interested in knowing as to in which situation do we need to use one of the above numbers and in which other situation we would use the other number.

From many of the answers I now understand that bot of the above numbers are complex numbers but they are of different types in a way that one has a single value while the other has more than 1 values. So in which practical situation do we prefer to use one form and in which the other?

• What is $2^i$? – José Carlos Santos Feb 1 '19 at 12:26
• @AlexSilva What is your $a$? If $a \neq e$ then $a^{yi} \neq \cos \, y+i\sin \, y$. – Kavi Rama Murthy Feb 1 '19 at 12:34
• Sorry, I mean $e$. – Alex Silva Feb 1 '19 at 12:36
• Do you know that $a^{x+iy} = e^{(x+iy)\ln a} = a^{x}\left(\cos (y\ln a) + i\sin(x \ln a)\right)$? – Alex Silva Feb 1 '19 at 12:42
• Would you have any trouble explaining the difference between $2\cdot 5$ and $2^5$? Your problem is the same, really. – Arthur Feb 1 '19 at 12:49

After thinking a lot about what was actually being asked (and your comments), I think I might have an answer explaining the differences between $$2i$$ and $$2^i$$.

(For the benefit of those unfamiliar with complex numbers, the expression $$2i$$ represents the complex number $$\sqrt {-2}$$, but it can also be represented as $$0+2i$$ in $$a+bi$$ (rectangular) form...as $$2 \angle \frac {\pi}{2}$$ in polar form (using $$|z| \angle \arg (z)$$) and as $$2e^{\frac {i\pi}{2}}$$ in exponential form (using $$|z|e^{i \arg (z)}$$) - in polar and exponential form, $$|z| = \sqrt {a^2+b^2}$$ and $$\arg (z) = \arctan \frac {b}{a}$$.)

We use the rectangular form of $$2i$$ when we have roots for a quadratic equation that have a negative discriminant $$\Delta = b^2 - 4ac <0)$$. (Rather than writing $$\sqrt{-1}$$ all the time, we just use $$i$$.)

We use the polar or exponential form to show the ease of multiplying, dividing, raising to powers or taking roots versus doing so in rectangular form. We can also express complex numbers in polar or exponential form for alternating-current circuits to show voltage lead or lag. In this case, $$2i$$ represented as $$2 \angle \pi/2$$ means the voltage is leading by $$\pi/2$$ radians.

The expression $$2^i$$ is a multi-valued function whose general value is $$2^i = e^{i \ln 2 + 2\pi k} =\cos (\ln 2 +2\pi k) + i \sin (\ln 2 + 2\pi k), k \in \mathbb {Z}.$$ The principal value ($$k=0$$) is $$2^i = e^{i \ln 2} =\cos (\ln 2) + i \sin (\ln 2)$$

You would probably use $$2^i$$ to illustrate the multiple values of a complex exponent and their relationship to natural logarithms. There is rotation involved when we have multiple values of $$k$$; for $$k>0$$ the rectangular values can vary extensively, but often we consider the principal value at $$k=0.$$

• Hmmm... But $2=e^{\log2+42i\pi}$ hence $2^i=e^{i\log2-42\pi}=e^{-42\pi}\cos(\log2)+ie^{-42\pi}\sin(\ln2)$, right? – Did Feb 1 '19 at 14:49
• Yes, you are correct. I was leaning towards the principal value of $2^i$ but $2^i$ is indeed a multi-valued function - it would depend on how in-depth the explanation the poster wants to give. I've corrected my post to indicate this. – bjcolby15 Feb 1 '19 at 20:28
• +1 for some clear and concise explanations. – Did Feb 1 '19 at 21:00
• Can you explain the difference in terms of simple rotation and rotation on unit circle in radians? – scico111 Feb 6 '19 at 9:47
• I feel that complex numbers are different from real numbers that complex numbers has to do something related to rotation.. that's why we see sine and cosine terms in the expanded forms of real numbers. So if this perspective is correct then how can we explain the difference between above two numbers from the point of view of how these numbers represent two different forms of rotations? – scico111 Feb 7 '19 at 10:04

$$2^i=\mathrm e^{i\ln 2}=\cos(\ln 2)+i\sin(\ln 2)\ne 2i.$$

• Hmmm... But $2=e^{\log2+42i\pi}$ hence $2^i=e^{i\log2-42\pi}=e^{-42\pi}\cos(\log2)+ie^{-42\pi}\sin(\ln2)$, right? – Did Feb 1 '19 at 14:49
• Yes but the definition for a positive base is the one I used, conventionally. See, for instance Wikipedia, Complex_exponents_with_positive_real_bases. – Bernard Feb 1 '19 at 15:11
• Interesting. I wonder who wrote this paragraph of the WP page, unfortunately with no supporting reference. – Did Feb 1 '19 at 17:16

$$2^{x+iy}=(e^{\ln(2)})^{(x+iy)}=e^{x\ln(2)+i\cdot y\ln(2)}=2^x\cdot e^{i \cdot y\ln(2)}=2^x(\cos {(y\ln2)}+i\cdot \sin {(y\ln2)})$$
$$2^{x+iy}=2^x[\cos(\ln(2y))+i\cdot \sin(\ln(2y))]$$
Here, $$2^i=2^{0+1i}$$
So, $$(x,y)=(0,1)$$

It becomes: $$2^i=2^0[\cos(\ln(2)+i\sin(\ln(2)]=\cos(\ln2)+i\cdot \sin(\ln2)$$

• Hmmm... But $2=e^{\log2+42i\pi}$ hence $2^i=e^{i\log2-42\pi}=e^{-42\pi}\cos(\log2)+ie^{-42\pi}\sin(\ln2)$, right? – Did Feb 1 '19 at 14:50