Complex Numbers: find minimum value of |z1 + a*z2| for real a

[rakrak1998]

Given the complex numbers z₁ = 1 + 3i and z₂ = -1 - i.

(a) Write down the exact values of |z₁| and arg(z₂).

(b) Find the minimum value of \(\displaystyle \, \lvert\, z_1\, +\, \alpha\, z_2\, \rvert ,\, \) where \(\displaystyle \, \alpha\, \in\, \mathbb{R}.\)

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I understand part A, now for part B i am relativity confused. I am not entirely sure how to find the modulus for 2 complex numbers multiplied with a value a.

Here is where i am at,

I tried to approach it like a regular modulus for a complex number, so took the modulus for z1, squared it, and then the modulus for z2, and squared that as well.

sqrt((sqrt(10))²+a²*(sqrt(2))²)

This does not match the answer at all. Any help?

 

[Prove It]

I understand part A, now for part B i am relativity confused. I am not entirely sure how to find the modulus for 2 complex numbers multiplied with a value a.

Here is where i am at,

I tried to approach it like a regular modulus for a complex number, so took the modulus for z1, squared it, and then the modulus for z2, and squared that as well.

sqrt((sqrt(10))²+a²*(sqrt(2))²)

This does not match the answer at all. Any help?

You might be able to get a lower bound using the reverse triangle inequality.

 

[pka]

I understand part A, now for part B i am relativity confused. I am not entirely sure how to find the modulus for 2 complex numbers multiplied with a value a.

Surely you can minimize \(\displaystyle \sqrt{(1-\alpha)^2+(3-\alpha)^2}~?\)

 

[Deleted member 4993]

I understand part A, now for part B i am relativity confused. I am not entirely sure how to find the modulus for 2 complex numbers multiplied with a value a.

Here is where i am at,

I tried to approach it like a regular modulus for a complex number, so took the modulus for z1, squared it, and then the modulus for z2, and squared that as well.

sqrt((sqrt(10))²+a²*(sqrt(2))²)

This does not match the answer at all. Any help?

Z₂ = -1+ (-1)i → αZ₂ = (-1)α+ (-1)αi

Z₁ = 1+ 3i

Z₁ + αZ₂ = 1+ 3i + (-α - αi) = (1 - α) + (3 - α)i

|Z₁ + αZ₂| = ??

 

[rakrak1998]

Z₂ = -1+ (-1)i → αZ₂ = (-1)α+ (-1)αi

Z₁ = 1+ 3i

Z₁ + αZ₂ = 1+ 3i + (-α - αi) = (1 - α) + (3 - α)i

|Z₁ + αZ₂| = ??

I Understand now! Thank you!

 

[Brady]

Z₂ = -1+ (-1)i → αZ₂ = (-1)α+ (-1)αi

Z₁ = 1+ 3i

Z₁ + αZ₂ = 1+ 3i + (-α - αi) = (1 - α) + (3 - α)i

|Z₁ + αZ₂| = ??

Hi there, is there anyway you could explain the last line?

How does one get from Z₁ + aZ₁ = 1+ 3i + (-α - αi) = (1 - α) + (3 - α)i, to the next equation and solve?

 

[Brady]

I don't understand how to proceed from Z₁ + aZ₂ =... --> to |Z₁ + aZ<sub>2| =...

Could you please elaborate?</sub>

 

[Brady]

Could you please explain further? I don't quite understand..

 

[stapel]

Could you please explain further? I don't quite understand..

What did you get when you worked from what the helper gave you? Using the parts and pieces, what did you come up with?

Please be complete. Thank you!

 

[pka]

Could you please explain further? I don't quite understand..

If \(\displaystyle y=\sqrt{(1-\alpha)^2+(3-\alpha)^2}~?\)
Then \(\displaystyle y'=\dfrac{-(1-\alpha)-(3-\alpha)}{\sqrt{(1-\alpha)^2+(3-\alpha)^2}}\)
How is \(\displaystyle y'=0~?\)

 

[Deleted member 4993]

Hi there, is there anyway you could explain the last line?

How does one get from Z₁ + aZ₁ = 1+ 3i + (-α - αi) = (1 - α) + (3 - α)i, to the next equation and solve?

First of all - it is NOT "aZ₁" it is "αZ₂".

What do you get - when you multiply 'Z₂' by 'α'?