Author Topic: Complex number inequality question (Read 1401 times) Tweet Share

006896 · « **on:** December 19, 2018, 11:02:41 am »

Hi friends,
Could you help me with the attached question? It's from Cambridge 4U textbook, Chapter 2.3. I've seen the textbook worked
solutions, but they don't make sense and I don't think I'd ever be able to replicate the answer in a test.
Thanks
EDIT: Sorry, I should've mentioned earlier (I didn't realise that the cropped area of the photo was so large) that the question I need help with is Q6.

RuiAce · « **Reply #1 on:** December 19, 2018, 11:23:15 am »

The solutions seem fine to me.
$\text{Firstly, it is }\textbf{expected}\text{ that you know the triangle inequality}\\ \boxed{|z_1+z_2|\leq |z_1|+|z_2| }.\\ \text{You should have seen this in your studies.}$
If not, then something's wrong.
$\text{You are also expected to know the condition for equality to hold.}\\ \text{The condition }|z_1+z_2| = |z_1| + |z_2|\text{ occurs}\\ \text{when }\boxed{\arg z_1 = \arg z_2},\text{ or equivalently they lie on the same ray from the origin.}$
$\text{This is where the condition }z_2 = kz_1\text{ comes from, where }k\text{ is a positive real number.}$
The idea is that two complex arguments can only have the same argument, if one is the other times a positive real number. If you multiply it with a negative real number, the argument becomes its negative as well. And if you multiply it with another complex number, the argument completely changes into something else.
$\text{So given }|z_2| = 6\text{, this is why we can now sub to obtain}\\ \boxed{|kz_1| = 6}.\text{ But of course, this just becomes }\\ k|z_1| = 6\text{, since }k\text{ is a positive real number here.}$
Lastly they finish it off by manually computing $|z_1| = |24+7i| = \sqrt{24^2+7^2} = 25$ to find the value of $k$ required, which consequently lets you find $z_2$.

The other one is done similarly, but it uses the reverse triangle inequality instead.

006896 · « **Reply #2 on:** December 19, 2018, 11:38:13 am »

Quote from: RuiAce on December 19, 2018, 11:23:15 am

The solutions seem fine to me.
$\text{Firstly, it is }\textbf{expected}\text{ that you know the triangle inequality}\\ \boxed{|z_1+z_2|\leq |z_1|+|z_2| }.\\ \text{You should have seen this in your studies.}$
If not, then something's wrong.
$\text{You are also expected to know the condition for equality to hold.}\\ \text{The condition }|z_1+z_2| = |z_1| + |z_2|\text{ occurs}\\ \text{when }\boxed{\arg z_1 = \arg z_2},\text{ or equivalently they lie on the same ray from the origin.}$
$\text{This is where the condition }z_2 = kz_1\text{ comes from, where }k\text{ is a positive real number.}$
The idea is that two complex arguments can only have the same argument, if one is the other times a positive real number. If you multiply it with a negative real number, the argument becomes its negative as well. And if you multiply it with another complex number, the argument completely changes into something else.
$\text{So given }|z_2| = 6\text{, this is why we can now sub to obtain}\\ \boxed{|kz_1| = 6}.\text{ But of course, this just becomes }\\ k|z_1| = 6\text{, since }k\text{ is a positive real number here.}$
Lastly they finish it off by manually computing $|z_1| = |24+7i| = \sqrt{24^2+7^2} = 25$ to find the value of $k$ required, which consequently lets you find $z_2$.

The other one is done similarly, but it uses the reverse triangle inequality instead.

Thanks so much for your reply, but I meant Q6. Sorry, I should have specified but I didn't realise that the photo had other questions in it.

RuiAce · « **Reply #3 on:** December 19, 2018, 11:59:26 am »

Quote from: 006896 on December 19, 2018, 11:38:13 am

Thanks so much for your reply, but I meant Q6. Sorry, I should have specified but I didn't realise that the photo had other questions in it.

That one really falls out of the statement that "the difference of (the lengths of) two sides of a rectangle" can never exceed (the length of) the third side. That kind of reasoning is all they really look for in the HSC.

The proof in the solutions is a more formal proof. In the HSC, if they want a formal approach, guidance will be provided.

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