# Complex Numbers (1 Viewer)

#### twinklegal19

##### .
What's the fastest way to simplify

$\bg_white \left ( 1+\cos{\frac{\pi}{8}}+i\sin{\frac{\pi}{8}}\right )^4+\left ( 1+\cos{\frac{\pi}{8}}-i\sin{\frac{\pi}{8}}\right )^4$

without expanding everything/using exact ratios?

Thanks!

#### deswa1

##### Well-Known Member
Use double anges to eliminate the 1's. For example, cos(pi/8) becomes 2cos(pi/16)^2 -1 which will 'delete' the 1. Then use double angles on the sin(pi/8)=2sin(pi/16)cos(pi/16). Factorise and then De Moivre's- BOOOOOM.

Nice question though

#### theind1996

##### Active Member
Use double anges to eliminate the 1's. For example, cos(pi/8) becomes 2cos(pi/16)^2 -1 which will 'delete' the 1. Then use double angles on the sin(pi/8)=2sin(pi/16)cos(pi/16). Factorise and then De Moivre's- BOOOOOM.

Nice question though
Haha fuark,

Strong ability/10.

Did not notice that.

#### twinklegal19

##### .
Thank you

Never thought of using double angles haha

#### deswa1

##### Well-Known Member
Haha fuark,

Strong ability/10.

Did not notice that.
Yeah I didn't know how to do that until there was a question in Terry Lee and that's where I learnt that wherever you have 1's in a trig equation, you can just 'delete' them like that. Its a nice trick

Or by geometry

#### theind1996

##### Active Member
Yeah I didn't know how to do that until there was a question in Terry Lee and that's where I learnt that wherever you have 1's in a trig equation, you can just 'delete' them like that. Its a nice trick
Does Terry have any other nice tricks in his book?

#### twinklegal19

##### .
Or by geometry
would you like to elaborate please?

I just solved it but go on!

#### jyu

##### Member
Find where 1+z and 1+z'(conjugate) are on the Argand diagram.

#### iSplicer

##### Well-Known Member
Just an addition: I was taught that, WHENEVER you see (1 + cos[something]) it should set off huge alarm bells. It's VERY likely that you need to use double angles to proceed =). Also, really helps to revise sum to products, and products to sum on the side (for those trickier 4u questions).

#### SpiralFlex

##### Well-Known Member
This method is long, but what I have done is used the identity repeatedly of

$\bg_white (a+b)^2-2ab=a^2+b^2$

This may not help with this question but it will certainly help you in the future. Good to find alternate ways than to use the traditional textbook methods.

$\bg_white Let\; z=(1+\cos \frac{\pi}{8}+i\sin\frac{\pi}{8}), \; \overline{z}=(1+\cos \frac{\pi}{8}-i\sin\frac{\pi}{8})$

The expression becomes,

$\bg_white z^4+(\overline{z})^4=(z^2+(\overline{z})^2)^2-2(z\overline{z})^2$

$\bg_white =((z+\overline{z})^2-2z\overline{z})^2-2(z\overline{z})^2$

$\bg_white =((2Re(z))^2-2|z|^2)^2-2|z|^4$

#### Nooblet94

Does Terry have any other nice tricks in his book?

You'll get it from school when we sign out.

#### theind1996

##### Active Member
You'll get it from school when we sign out.
Ah ok, that's good. Hurry the eff up and sign out

#### twinklegal19

##### .
Just an addition: I was taught that, WHENEVER you see (1 + cos[something]) it should set off huge alarm bells. It's VERY likely that you need to use double angles to proceed =). Also, really helps to revise sum to products, and products to sum on the side (for those trickier 4u questions).
Thanks! Just realised how it can also eliminate -1 with double angles as well. Will be sure to remember that next time haha

This method is long, but what I have done is used the identity repeatedly of

$\bg_white (a+b)^2-2ab=a^2+b^2$

This may not help with this question but it will certainly help you in the future. Good to find alternate ways than to use the traditional textbook methods.

$\bg_white Let\; z=(1+\cos \frac{\pi}{8}+i\sin\frac{\pi}{8}), \; \overline{z}=(1+\cos \frac{\pi}{8}-i\sin\frac{\pi}{8})$

The expression becomes,

$\bg_white z^4+(\overline{z})^4=(z^2+(\overline{z})^2)^2-2(z\overline{z})^2$

$\bg_white =((z+\overline{z})^2-2z\overline{z})^2-2(z\overline{z})^2$

$\bg_white =((2Re(z))^2-2|z|^2)^2-2|z|^4$
Thanks I'm glad to learn any alternative methods, will definitely try to remember this for future problems.