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If I have |z|^n = |1| and |(z+1)|^n = |1|, can I take the nth root of both sides to say that |z|=|z+1|=1?

 Mar 16, 2023
 #1
avatar+195 
+1

Not necessarily. 

If |z|^n = |1|, then we know that |z| = 1 because the absolute value of any complex number raised to a positive power (such as n) is always non-negative. 

However, if |(z+1)|^n = |1|, taking the nth root of both sides only tells us that |z+1|^(1/n) = 1. It does not necessarily imply that |z+1| = 1. 

For example, consider z = -1/2 - i√3/2. We have |z|^n = |1| since |z| = 1, but also |(z+1)|^n = |1| since |z+1| = |1/2 - i√3/2| = 1. However, |z| ≠ 1 and |z+1| ≠ 1. 

So in general, we cannot conclude that |z| = |z+1| = 1 from the given information.

 Mar 16, 2023
 #2
avatar+132 
+1

So if I have z^n = (z + 1)^n = 1, and spilt it into two equations: z^n = 1 and (z+1)^n = 1, and take their magnitudes, is that enough to say |z|=|z+1|=1 because it must be a root of unity? 

Saphia1123  Mar 16, 2023
 #3
avatar+195 
+1

Yes, you are correct! If z^n = (z + 1)^n = 1, then we can split it into two equations: z^n = 1 and (z+1)^n = 1. Taking the magnitudes of both sides of these equations, we get:

|z|^n = 1 and |z+1|^n = 1

Since the magnitudes of complex numbers are always non-negative, these equations imply that |z| = 1 and |z+1| = 1. 

Furthermore, since z^n = 1, we know that z must be a root of unity (a complex number of unit magnitude that satisfies z^n = 1). Since |z| = 1, z lies on the unit circle in the complex plane. 

Similarly, since (z+1)^n = 1, we know that z+1 must also be a root of unity. Since |z+1| = 1, z+1 also lies on the unit circle in the complex plane. 

Therefore, we can conclude that |z| = |z+1| = 1, and z and z+1 are both roots of unity lying on the unit circle in the complex plane.

Justingavriel1233  Mar 16, 2023
 #4
avatar+132 
0

How could I find the possible values of z in exponential form?  For |z|^n=1, could I just say:

the exponential form for |z|^n = 1 is e^{\frac{2\pi i}{n}} for n^{th} roots of unity.

 

 

Idk how to approach finding the exponential form for z+1

Saphia1123  Mar 16, 2023
edited by Saphia1123  Mar 16, 2023
edited by Saphia1123  Mar 16, 2023

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