- #1

Taleb

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- MHB
- Thread starter Taleb
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In summary, for $u = a+bi$ and $v = c+di$, if $|u|$ and $|v|$ are both less than 1, then $a^2+b^2<1$ and $c^2+d^2<1$. When $u+v$ is calculated, it can be shown that $|u+v| < \sqrt{2+2(ac+bd)}$. It can also be proven by induction that $|u-v| \leq \sqrt{2-2(ac+bd)}$.

- #1

Taleb

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- #2

HOI

- 921

- 2

u+ v= (a+ c)+(b+ d)i. $|u+v|= \sqrt{(a+ c)^2+ (b+ d)^2}=$$\sqrt{a^2+ 2ac+ c^2+ b^2+ 2bd+ d^2}= $$ \sqrt{(a^2+ b^2)+ (c^2+ d^2)+ (2ac+2bd)}< \sqrt{1+ 1+ 2(ac+ bd)}< \sqrt{2+ 2(ac+ bs)}$

Can you prove that $ac+ bd$ is less than 1/2?

- #3

Opalg

Gold Member

MHB

- 2,778

- 13

Problem 2) seems to be a lot harder. I found a sketch here of how to prove it by induction (again with $\sqrt2$ rather than $\sqrt3$).

Complex numbers are numbers that have both a real part and an imaginary part. They are represented in the form a + bi, where a is the real part and bi is the imaginary part with i being the square root of -1.

The modulus of a complex number is its distance from the origin on the complex plane. When the modulus is less than or equal to 1, it means that the number is within a circle with a radius of 1 centered at the origin.

Complex numbers with modulus less than or equal to 1 are represented within a unit circle on the complex plane. The center of the circle is at the origin, and the numbers are located at various points along the circumference of the circle.

Complex numbers with modulus less than or equal to 1 have many applications in mathematics, including in geometry, physics, and engineering. They are also used in solving equations and in signal processing.

Operations on complex numbers with modulus less than or equal to 1 are performed in the same way as operations on regular complex numbers. The only difference is that the result must also have a modulus less than or equal to 1.

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