Imaginary number question

[dhruv]

if z = 3+i, find the values of n for which Im(z^n)=0.

I got to:
18.435n = k*pi

 

[dhruv]

The answer is: n = [5,15,25,35....] but I don't know how to get to this.

 

[pka]

if z = 3+i, find the values of n for which Im(z^n)=0.
I got to: 18.435n = k*pi

The answer is: n = [5,15,25,35....] but I don't know how to get to this.

That answer is not correct. Who or what gave it to you?
The complex number \(\displaystyle z=3+i\) has argument \(\displaystyle \theta=\arctan\left(\frac{1}{3}\right)\).
Therefore, in order for \(\displaystyle \Im{z}^n=0\) it must be the case that that \(\displaystyle n=\frac{k\pi}{\theta}\text{ for }k\in\mathbb{Z}\)

 

[Jagulba]

I got to sin(k*pi/n)=(1/10)^0.5
but I don't know how you would find or know that sin(pi/5) is (1/10)^0.5

 

[Dr.Peterson]

if z = 3+i, find the values of n for which Im(z^n)=0.

I got to:
18.435n = k*pi

What you did is correct, and agrees with what pka showed.

The answer is: n = [5,15,25,35....] but I don't know how to get to this.

The claimed answer (which I am guessing is supposed to be the set {5, 15, 25, 35, ...}) clearly is not correct: https://www.wolframalpha.com/input/?i=(3+i)^5

That shows that z^5 is not pure real.

 

[dhruv]

That answer is not correct. Who or what gave it to you?
The complex number \(\displaystyle z=3+i\) has argument \(\displaystyle \theta=\arctan\left(\frac{1}{3}\right)\).
Therefore, in order for \(\displaystyle \Im{z}^n=0\) it must be the case that that \(\displaystyle n=\frac{k\pi}{\theta}\text{ for }k\in\mathbb{Z}\)

I completely agree with you. The answer key that I provided is in my Oxford mathematics book....I don't know how can it be so off.

 

[Jagulba]

if z = 3+i, find the values of n for which Im(z^n)=0.

N[solve(pi/x-pi^3/(6*x^3)==(1/10)^0.5)] - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

OK i went for expansion of Sinx=x-x³/3!
I went for k=1 and got to roots of it, and n seems to be close to 10 ... still I think it needs more of terms in the expansion to get to zero though

This one is change a bit in last digits to get close to zero (the solution I got 9.76315 and I changed it to 9.7640629)

(3+i)^(9.764062903) - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

not sure though because not all of the roots give zero, so not sure what to think of it

 

[dhruv]

What you did is correct, and agrees with what pka showed.

The claimed answer (which I am guessing is supposed to be the set {5, 15, 25, 35, ...}) clearly is not correct: https://www.wolframalpha.com/input/?i=(3+i)^5

That shows that z^5 is not pure real.

Thanks! So does that mean that the first possible value of n is: 9.764
as:
pi / tan^-1(1/3) = 9.764

 

[Dr.Peterson]

Yes.

 

[Jagulba]

exactly:

(3+i)^(pi/arctan(1/3)) - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com