- #1
Taleb
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Complex numbers such that modulus (absolute value) less than or equal to 1.
- 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 $|u-v| \leq \sqrt{2-2(ac+bd)}$.
- #2
HOI
- 921
- 2
Write u= a+ bi and v= c+ di. If modulus u and v are both less than 1 the $\sqrt{a^2+ b^2}< 1$ and $\sqrt{c^2+d^2}< 1$ so $a^2+ b^2< 1$ and $c^2+ d^2< 1$.
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?
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
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Following up on Country Boy's calculation, notice that if $v$ is replaced by $-v$ then $b$ becomes $-b$ and $d$ becomes $-d$. Therefore $$|u+v| \leqslant \sqrt{2+2(ac+bd)}, \qquad |u-v| \leqslant \sqrt{2-2(ac+bd)}.$$ It follows that if $ac+bd>0$ then $|u-v| \leqslant\sqrt2$, and if $ac+bd<0$ then $|u+v|\leqslant\sqrt2$. That proves 1). (In fact it proves a stronger result, with $\sqrt2$ instead of $\sqrt3$.)
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 $\sqrt3$).
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 $\sqrt3$).
FAQ: Complex numbers such that modulus (absolute value) less than or equal to 1.
What are complex numbers?
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.
What does it mean for a complex number to have a modulus less than or equal to 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.
How are complex numbers with modulus less than or equal to 1 represented on a complex plane?
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.
What is the significance of complex numbers with modulus less than or equal to 1 in mathematics?
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.
How are operations performed on complex numbers with modulus less than or equal to 1?
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.