If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|.
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 …
$|z+w|=|z|+|w|$ iff $z=cw$ - Mathematics Stack Exchange
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complex numbers - If $z,w \in {\mathbb C}\setminus\{0\}$, prove …
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absolute value - If $z$ and $w$ are complex numbers can we use …
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 …
complex numbers - Show that $|z - w| \geq \big||z|-|w|\big ...
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, …
Knowing $|z|$ and $|w|$ (and possibly $z\\overline{w}$), what are …
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 …
Show that $|z + w|^2 = |z|^2 + |w|^2 + 2\text {Re} (z\bar w)$ for …
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Prove if $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over …
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 …
If $z$ and $w$ are complex numbers such that $| (z+w)| = | (z-w ...
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.
Show that $|z+w|^2$ + $|z-w|^2$ = $2|z|^2 - Mathematics Stack …
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 …