1. ## Complex numbers Questions

Hi,

I have recently started doing a further maths "a level" course, and im having a bit of diffuculty with some of the harder complex numbr questions.

The last two i have been working on for hours, but I just cant seem to get them, its probably something simple but anyway;

1. given that a and b are real constants, prove that "ai" is a root of the equation;

z³-bz² +a²z- a²b=0

and find the roots of the equation in terms of a and b.

2. Simplify without the use of a calculator:

(cos π/7 -i sin π/7)/(cos π/7 + i sin π/7)

I have tried these many times but to no avail. Thanks for your help,

Owen.

2. Hi,

I have recently started doing a further maths "a level" course, and im having a bit of diffuculty with some of the harder complex numbr questions.

The last two i have been working on for hours, but I just cant seem to get them, its probably something simple but anyway;

1. given that a and b are real constants, prove that "ai" is a root of the equation;

z³-bz² +a²z- a²b=0

and find the roots of the equation in terms of a and b.

2. Simplify without the use of a calculator:

(cos π/7 -i sin π/7)/(cos π/7 + i sin π/7)

I have tried these many times but to no avail. Thanks for your help,

Owen.
hi
if you have to prove that "ai" is a root of the equation;
z³-bz² +a²z- a²b=0

then simply put z=ai
and see if L.H.S=R.H.S

since in the first part it is already given that ai is one of the root of the equation
therefore from the property of the root of the equation that complex numbers occur in pair, complex number and its conjugate,therefore -ai is also one of the root
now, since the last term in the equation is the product of all the roots
therefore, our third root is -b.

in second question use this property
(cosQ + i*sinQ)= (e^(iQ))
(cosQ - i*sinQ)= (e^(-iQ))

3. ## Thanks

Haha thank you I now get the first one: I am an idiot

For the second I have never seen that property :S

(cosQ + i*sinQ)= (e^(iQ))
(cosQ - i*sinQ)= (e^(-iQ))

Unless its written in a different way than i'm used to or something like that...

What is e?