1. ## complex numbers

Express the comlex number (√3+i) in trigonometric form. Hence find the smallest positive integer n that (√3+i)^n is a real number.

This is how I have started it however im really not sure if this is right hoping someone could help.

ArgZ = tan^-1 (1/√3) = 30

|Z| = √3+1 = 2

Z = 2(cos30+isin30)

Is this correct??

Also how do i complete the question??

Thanks

2. To finish this problem it is best to use number notation.
$\displaystyle \left( {\sqrt 3 + i} \right) = 2\left( {\cos \left( {\frac{\pi } {6}} \right) + i\sin \left( {\frac{\pi }{6}} \right)} \right)$.
Now real numbers have arguments that are integral multiples of pi
So the second answer is n=6.

3. Using DeMoivre,

So you're finding n so that $\displaystyle 2^n(cos (nx) + i sin (nx))$ is real.

The only way to do that is if your imaginary part is zero, as in $\displaystyle sin (nx) = 0$.

Since x is 30 degrees, your first option is n=6, as $\displaystyle sin (6*30) = sin (180) = 0$.

4. Originally Posted by Plato
Now real numbers have arguments that are integral multiples of pi
Could you please explain this sentence lol, unsure where the 6 came from :/ thanks

5. still on the topic of complex numbers, could anybody tell me if this is correct?

complex number Z and conjugate Zbar satisfy equation

3Z + Zbar = (11+7i)/(1+i)

find Z in form x + iy

This is how i attempt to answer it but unsure if its correct

let Z = x + iy
Zbar = x - iy

3(x+iy) + (x-iy) = (11+7i)/(1+i)

3 (x+iy) + (x-iy) = 10 + 6i

2(x + iy) = 10 + 6i

(x+iy) = (10 + 6i))/2

x + iy = 5 + 3i

$\displaystyle \frac{{11 + 7i}}{{1 + i}} = \frac{{\left( {11 + 7i} \right)\left( {1 - i} \right)}} {2} = 9 - 4i$.
$\displaystyle \begin{gathered} 3z + \overline z = 9 - 2i \hfill \\ 3(x + yi) + (x - yi) = 9 - 2i \hfill \\ 4x + 2yi = 9 - 2i \Rightarrow \quad x = \frac{9}{4}\,\& \,y = - 1 \hfill \\ \end{gathered}$