Help with Cambridge Extension 2 Enrichment Question - Complex Numbers (1 Viewer)

Poopygerald · Dec 22, 2022

Hello,

I am having trouble doing q a(iii) and b. Could someone help me please? Thanks.

idkkdi · Dec 22, 2022

Poopygerald said:
View attachment 37262

Hello,

I am having trouble doing q a(iii) and b. Could someone help me please? Thanks.

a) iii) let tan theta = 1/3
b) a) iii) tells u stuff about angles of the fourth roots, do sth and then also slap on a root5 for modulus

Poopygerald · Dec 23, 2022

idkkdi said:
a) iii) let tan theta = 1/3
b) a) iii) tells u stuff about angles of the fourth roots, do sth and then also slap on a root5 for modulus

Sorry I'm not sure how to use your advice for a) iii). I would be glad if you could send your working. Thanks.

idkkdi · Dec 23, 2022

Poopygerald said:
Sorry I'm not sure how to use your advice for a) iii). I would be glad if you could send your working. Thanks.

let t be theta.

4t = 4arctan(1/3) = arctan ( ..... [sub in tant = 1/3 into that expression])

5uckerberg · Dec 23, 2022

Poopygerald said:
Sorry I'm not sure how to use your advice for a) iii). I would be glad if you could send your working. Thanks.

By letting $\tan{\theta}=\frac{1}{3}$ you are stating that $\theta=\tan^{-1}\frac{1}{3}$ which is exactly what you want considering where you finished from part ii. Considering the question is stating that $\tan^{-1}\frac{24}{7}=4\tan^{-1}\frac{1}{3}$ by using $\theta=\tan^{-1}\frac{1}{3}$ we then have $\tan^{-1}\frac{24}{7}=4\theta$ leading to $\frac{24}{7}=\tan{4\theta}$ .

Using $\tan{\theta}=\frac{1}{3}$ and part ii $\tan{4\theta}=\frac{4\times{\frac{1}{3}}-4\times{\left(\frac{1}{3}\right)^{3}}}{1-6\times{\left(\frac{1}{3}\right)^{2}}+6\times{\left(\frac{1}{3}\right)^{4}}}$

b) For the four fourth roots of $7+24i$ the mod arg form is $25\left(\cos\left(\tan^{-1}\frac{24}{7}\right)+i\sin\left(\tan^{-1}\frac{24}{7}\right)\right)$
using part iii
$25\left(\cos\left(\tan^{-1}\frac{24}{7}\right)+i\sin\left(\tan^{-1}\frac{24}{7}\right)\right)=25\left(\cos\left(4\tan^{-1}\frac{1}{3}\right)+i\sin\left(4\tan^{-1}\frac{1}{3}\right)\right)$ , then considering this pattern is repeated it becomes $25\left(\cos\left(4\tan^{-1}\frac{1}{3}+2n\pi\right)+i\sin\left(4\tan^{-1}\frac{1}{3}+2n\pi\right)\right)$ .

Find the 4th root of that, and then use cosine and sine trig expansions to finish the question

Poopygerald · Dec 25, 2022

I appreciate the help from you guys, you've really saved my brain from frying. Thanks again for your time.

Make a Difference – Donate to Bored of Studies!

Help with Cambridge Extension 2 Enrichment Question - Complex Numbers (1 Viewer)

Poopygerald

New Member

idkkdi

Well-Known Member

Poopygerald

New Member

idkkdi

Well-Known Member

5uckerberg

Well-Known Member

Poopygerald

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)