# What is the benefit of representing a complex number as e^i(theta) versus e^(a+bi), what is the process of finding a solution to this example?

What is the benefit of representing a complex number as $e^{i\theta}$ versus $e^{a+bi}$? Am I correct in saying that these give the same information but offer convenience in different situations?

Use of $e^{a+bi}$ example:

Find all complex numbers z = a + bi such that $e^z = -2$

The solution and process for arriving at the solution is

$e^{a+bi} = e^{a}e^{bi}$ so $\lvert e^{a+bi} \rvert = e^a.$ Since $\lvert -2 \rvert = 2, a = ln 2.$ and $Arg(e^{a+bi}) = b$ up to adding multiples of $2\pi Arg(-1) = \pi,$ so b is any odd multiple of $\pi.$

i.e. Answer: $ln2 + bi\pi$

This is a problem from MIT OCW 18.03, PSII.6(b)

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/assignments/MIT18_03S10_ps2s.pdf

• Do you mean $e^{\theta i}$ versus $a+bi$? Obviously it makes no difference whether you use the letter $b$ or $\theta$ for the imaginary part of the exponent. Jun 21, 2016 at 21:13
• Two big reasons: Euler's identity and the fact that $e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}$ Jun 21, 2016 at 21:14

Usually when one writes $e^{i\theta}$ the parameter $\theta$ is understood to be real. Assuming this there is a huge difference: you can't express every complex number as $e^{i\theta}$. The reason is that, due to Euler's identity, the complex number $e^{i\theta}$ must have module (length if you see $\mathbb{C}$ only as the vector space $\mathbb{R}^2$) 1. So only the points that lie in the unit circumference (in the complex plane) can be expressed as $e^{i\theta}$.
Now, let's look at the representation $e^{a+bi}$ (as before $a$ and $b$ are usually supposed to be real). The complex exponential satisfies by definition (motivated by Eulers indentity) that $e^{a+ib}=e^ae^{ib}$. We've seen that $e^{ib}$ is a point on the complex unit circumference. Multiplying it by $e^a$ corresponds just to change its module (again, seeing $\mathbb{C}$ as $\mathbb{R}^2$ this is to say that the vector direction does not change) to $e^a$. If you keep $a$ fixed (and move $b$) then the points $e^{a+bi}$ are just the complex circunference centered in the origin and with radius $e^a$. Any nonzero point in the complex plane is in one such circunference so, as the real exponential is a bijection from $\mathbb{R}$ to $(0,+\infty)$, it follows that every complex number except $z=0$ have the representation $e^{a+bi}$.
Notice that because of the real exponential being bijective (in the above sense) the representation $e^{a+bi}$ is the same that $re^{ib}$, where $r$ is a real positive number. This last one is called polar representation or polar form of complex numbers. This one is usually introduced after the so called vector representation of complex numbers $z=a+bi$, implicitly denoting that as a vector space $\mathbb{C}$ is nothing but $\mathbb{R}^2$. Notice also that so far we've just looked at the cartesian and polar coordinates in $\mathbb{R}^2$ from a complex setting. This is justified cause the striking property that makes the complex plane so interesting is that you can multiply complex numbers with the same nice algebraic properties that the real multiplication. In this case the relevant implication is that the complex exponential (besides the above identity) satisfies $e^{i(b_1+b_2)}=e^{ib_1}e^{ib_2}$. As a result, one can compute the multiplication of two complex numbers just by multiplying their modules and adding their arguments ("their angles"). So, besides the usual cartesian-polar issues, the conveniences for using the polar form of complex numbers is that it is the suitable form for multiplication, and the conveniences for using the vector form is that it is suitable for summation.
• To "see" that angles add must you use trigonometric identities and euler's identity $e^{i\theta} = cos(\theta\ + isin(\theta)$? I see how a function $e^{(a + bi)t}$ can generate a logarithmic spiral now. Jun 25, 2016 at 22:55
• @AnthonyOrona There are several ways to define the complex exponential. With each one of them one can verify that $e^{z+w}=e^ze^w$, and so the angles add when multiplying. With this definition motivated by Euler's identity the way of proving it is (with the properties of the real exponential and) the sum-angle-formulas for the sine and cosine functions. So yes, from this setting you mustuse the trigonometric identities to see that angles add when multiplying. Ironically this is usually used the other way around; this propert of complex exponential is a good mnemotecnic to trig identities. :)