<p>if z = (cos 30degrees + isin30degrees), find a value of n, n cannot equal 0, for which z^n = 1</p>
<p>i believe its true for all values of n because the magnetude of the equation is 1 (the coefficient in front of the cos/sin is 1) i think. Ive never seen it written like that tho.</p>
<p>I don’t think there is one. I could be way off, but:
(cos30d+isin30d) = (rt3/2+i/2)
(rt3/2+i/2)^n = 1
log base(rt3/2+i/2) 1 = n
log(1)/log(rt3/2+i/2) = n
Punch that into a calculator in a+bi mode and you’ll get 0. I tried solving it with de Moivre’s Theorem and ended up with 0 again.</p>
<p>yea i keep getting zero too…</p>
<p>it probably is not z^n = 1. The n is kind of fuzzy and you can’t really see it. What should it be if it’s not z^n = 1?</p>
<p>where did you get that question?</p>
<p>well it can’t be <1… you end up having to take natural log of 1 which = 0 and that’s what screws it up, maybe it’s 10?</p>
<p>lol seriously…where do you get these questions from?what class is it?</p>
<p>If such an n existed, then z would necessarily be a complex nth root of unity, meaning that there exists some regular polygon of which z is a vertex. One vertex of this polygon would be at the point (1,0), the others spaced uniformly about the unit circle. By the definition of complex nth roots of unity, n is the number of vertices. </p>
<p>The angular displacement between adjacent vertices would be 30 degrees. The number of vertices is then 360/30 = 12.</p>
<p>So n = 12.</p>
<p>
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<p>True, the absolute value of z raised to any power will be 1. However, by Euler’s formula, there exists an angle theta such that z = exp(i*theta). Now examine what happens:</p>
<p>z^n = [exp(i<em>theta)]^n = exp(i</em>n*theta)</p>
<p>The result is a new complex number of the same magnitude but a different angle. So in other words complex numbers not only can increase in magnitude when raised to a power–they also can rotate about the origin.</p>
<p>[By the way, in this case theta is trivially 30 degrees.]</p>
<p>formally backing up nilkn.</p>
<p>de Moivre’s formula
(cos(x) + i<em>sin(x))^n = cos(nx) + i</em>sin(nx)</p>
<p>mode: degrees
z = cos(30) + i<em>sin(30)
z^n = cos(30n) + i</em>sin(30n)
z^n = 1
cos(30n)=1 and sin(30n)=0
30n = 360k, where k is any integer
n = 12k
for k=1 n=12.</p>
<p>r3n-<em>-, if <n is=“” kind=“” of=“” fuzzy=“” and=“” you=“” can’t=“” really=“” see=“” it=“”>,
switch to o</n></em>o. :D</p>
<p>if this question showed up on ACT, i’d be 0_0. where do you get your questions? how many people asking you will it take to have you divulge your collegiate confidential sources? 
</p>