if z = 3+i, find the values of n for which Im(z^n)=0.
I got to: 18.435n = k*pi
That answer is not correct. Who or what gave it to you?The answer is: n = [5,15,25,35....] but I don't know how to get to this.
What you did is correct, and agrees with what pka showed.if z = 3+i, find the values of n for which Im(z^n)=0.
I got to:
18.435n = k*pi
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)^5The answer is: n = [5,15,25,35....] but I don't know how to get to this.
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.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 \(\displaystyle n=\frac{k\pi}{\theta}\text{ for }k\in\mathbb{Z}\)
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
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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)
not sure though because not all of the roots give zero, so not sure what to think of it(3+i)^(9.764062903) - Wolfram|Alpha
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Thanks! So does that mean that the first possible value of n is: 9.764What 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.