Question

Prove: $|z_1+z_2|\ge\frac{1}{2}(|z_1|+|z_2|)|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}|$

score 124 · Answer 1 · 2018-05-31 13:33:20

Prove: $|z_1+z_2|\ge\frac{1}{2}(|z_1|+|z_2|)|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}|$

Published on 31 May 2018 - 13:33

score 63 · Answer 2 · 2018-05-31 14:29:03

If $|z + \frac{1}{z}| = a$ where $a$ is a positive real number then why is $\frac{-a+\sqrt{a^2+4}}{2}\leq\lvert z\rvert\leq\frac{a+\sqrt{a^2+4}}{2}$?

score 64 · Answer 3 · 2018-05-31 14:33:49

If $\frac{a}{|z_2-z_3|}=\frac{b}{|z_3-z_1|}=\frac{c}{|z_1-z_2|}$, then why is $\frac{a^2}{z_2-z_3}+\frac{b^2}{z_3-z_1}+\frac{c^2}{z_1-z_2}=0$?

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159

Views

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289

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Let $w$ be a primitive root of a unit of order 3, prove that $(1-w+w^2)(1+w-w^2)=4$

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score 476 · Answer 12 · 2018-06-02 12:37:22

476

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How is equation of a circle in Argand plane given by $z\bar z + a \bar z + \bar a z+ b = 0$?

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score 75 · Answer 14 · 2018-06-02 23:56:27

75

Views

$(m,n)=1$, $a^m=1$, $b^n=1$, can we show $a+b+1\neq 0$?

Published on 02 Jun 2018 - 23:56

score 77 · Answer 15 · 2018-06-03 05:58:33

77

Views

How can $|z-(i \bar w)| = |z-(-i \bar w)|$ where z and w are complex numbers?

Published on 03 Jun 2018 - 5:58

List Question

Prove: $|z_1+z_2|\ge\frac{1}{2}(|z_1|+|z_2|)|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}|$

If $|z + \frac{1}{z}| = a$ where $a$ is a positive real number then why is $\frac{-a+\sqrt{a^2+4}}{2}\leq\lvert z\rvert\leq\frac{a+\sqrt{a^2+4}}{2}$?

If $\frac{a}{|z_2-z_3|}=\frac{b}{|z_3-z_1|}=\frac{c}{|z_1-z_2|}$, then why is $\frac{a^2}{z_2-z_3}+\frac{b^2}{z_3-z_1}+\frac{c^2}{z_1-z_2}=0$?

Complex number transformations

Let $w$ be a primitive root of a unit of order 3, prove that $(1-w+w^2)(1+w-w^2)=4$

Arguments of collinear complex numbers?

Reflection of a line $\bar az + a\bar z = 0$ in real axis?

Sketching a set in the Complex plane.

Find all $z \in \mathbb{C}$ such that $z^4 + c = 0$

Calculating residues of complex functions

What is the distribution of amplitude and phase of circularly symmetric normal random variable.

How is equation of a circle in Argand plane given by $z\bar z + a \bar z + \bar a z+ b = 0$?

Complex number absolute value inequality

$(m,n)=1$, $a^m=1$, $b^n=1$, can we show $a+b+1\neq 0$?

How can $|z-(i \bar w)| = |z-(-i \bar w)|$ where z and w are complex numbers?

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