complex numbers
Convert $e^z$ to $a+bi$
I'm having trouble figuring out this very simple problem. Below is my attempt, but can you really have $1/e$ as the modulus of a complex number?
$$z=-1+\frac{i\pi}{4}$$
$$e^{-1+\frac{i\pi}{4}}=e^{-1}e^{\frac{i\pi}{4}}$$
$$=\frac{1}{e}\cos(\pi/4)+i \sin(\pi/4)$$
$$=\frac{1}{e}\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)$$
The problem is that points can be expressed in polar form in more than one way. Take a look at the diagram below:
There's your point (in blue). As you can see, it's in the second quadrant, and the angle $\theta = \frac{2\pi}{3}$ passes through it. So one could accurately say that the point's polar representation is $r = 1, \theta = \frac{2\pi}{3}$, i.e. $$z = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)$$
You'll notice, though, that $\theta = -\frac{\pi}{3}$ represents the same angle, but in the opposite direction. With this in mind, one could just as accurately say that the point's polar representation is $r = -1, \theta = -\frac{\pi}{3}$, i.e. $$z = -\cos\left(-\frac{\pi}{3}\right) - i\sin\left(-\frac{\pi}{3}\right)$$
Because $\cos\left(\frac{2\pi}{3}\right) = -\cos\left(-\frac{\pi}{3}\right)$ and $\sin\left(\frac{2\pi}{3}\right) = -\sin\left(-\frac{\pi}{3}\right)$, these two polar representations are equivalent. In other words, there's nothing wrong with concluding, as you did, that $\theta = -\frac{\pi}{3}$ You and the so-called "required answer" can reasonably disagree here and both be correct.
But you still did something wrong. If you choose to use $\theta = -\frac{\pi}{3}$, then you must also choose $r = -1$, yet you choose $r = 1$, based on your modulus computation. How can you avoid making this mistake in the future?
Simple: look at the quadrant in which the point lies. The point $-\frac{1}{2} + \frac{\sqrt{3}}{2}i$ has negative real part and positive imaginary part. Therefore it must be in the second quadrant. The angle $\theta = -\frac{\pi}{3}$ extends into the fourth quadrant, so if you intend to use that angle, then you must also make $r$ negative.
$$z_1+z_2 = \cos(\pi/7)+2 +i (\sin(\pi/7)+1)$$
so
$$|z_1+z_2|^2 = (\cos(\pi/7)+2)^2 + (\sin(\pi/7)+1)^2 $$
so multiply out and use $\cos^2(\theta)+\sin^2(\theta)=1$ to get the desired result.
$|x+iy|$ is the modulus and for real $x$ and $y$ is $\sqrt{x^2+y^2}$.
Best Answer
You are right. The answer is correct. Modulus can be any positive value. So, $1\over e$ can surely be it.