Oct 14, 2016 · If z is given as x + iy, and w = u + iv, then |z| = √(x^2 + y^2), = r, say; and |w| = √(u^2 + v^2), = p, say Then we can also view z as r cis θ, and w as p cis φ (i.e., p (cos φ + i sin φ) in polar form; so that θ and φ can be determined from x, y, u, and v -- to within 2π.)

Nov 9, 2017 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Nov 11, 2015 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Feb 18, 2017 · Now, given $(z,w)\in\mathbb{C}^2$ : $$\vert z\vert=\vert(z-w)+w\vert\le\vert z-w\vert+\vert w\vert$$ Hence : $$\vert z\vert-\vert w\vert\le\vert z-w\vert$$ And by symmetry, we also have : $$\vert w\vert-\vert z\vert\le\vert w-z\vert$$ which can be written : $$-\left(\vert z\vert-\vert w\vert\right)\le\vert z-w\vert$$ Finally :

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Draw vectors on the complex plane from the origin to a point on the unit circle. Notice that the argument of the sum of two such vectors is the average of the arguments of the two vectors (observe that the latter is defined only modulo a multiple of $\pi$).

If z and w are complex numbers such that $|z+w|$ = $|z-w|$, prove that $\arg(z)-\arg(w)= \pm\pi/2$. Can this be solved algebraically or would a graphic interpretation be better.

You seem to be using $|z + w|^2 = a^2 + b^2 + c^2 + d^2$ and $|z - w|^2 = a^2 + b^2 - c^2 - d^2$. These are both false. Instead we have $$|z + w|^2 = |(a+bi) + (c+di)|^2 = |(a + c) + (b + d)i|^2 = (a + c)^2 + (b + d)^2$$ and a similar calculation gives $|z - w|^2 = (a-c)^2 + (b - d)^2$.