Getting the argand with De Moires theorem

In summary, the conversation discusses solving for the value of z^{\frac{1}{4}}, where z is a complex number in the second quadrant of the Argand diagram. The solution involves using De Moivre's Theorem and the oscillation of trigonometric functions. The argument must be greater than 1.1902 + 2k\pi for the angle to lie in the second quadrant. However, there is a discrepancy in the solution when considering the range of k, causing confusion.
  • #1
655
0
Hi just did an interesting question which made me wonder about solving it another way around

z = 2 + 5i
find the value of [/tex]z^{\frac{1}{4}}[/tex] which lies in the second quadrant of the Argand diagram and enter it's argument


So [tex]arctan(5/2) \approx 1.1902 [/tex]

We use De Moivre's Thereom and the the oscillation of the trigonometric functions
[tex]\sqrt{29}(cos(\frac{1.1902 + 2k\pi}{4}) + isin(\frac{1.1902 + 2k\pi}{4}))

So to lie in the second quadrant the angle must be [tex]\theta > \frac{\pi}{2}[/tex]
[tex]\theta > \frac{ \pi}{2}[/tex]
[tex]\frac{1.1902 + 2k\pi}{4} > \frac{\pi}{2}[/tex]

so k > 0.8 so k > 0 (where k is an integer)

The next integer after 0 is 1! So sub k = 1 into [tex]\frac{1.1902 + 2\pi}{4}[/tex]

then convert to degrees 107° agreed?

BUT how come if we say:
To lie in the second quadrant [tex]\theta < \pi[/tex]

[tex]\frac{1.1902 + 2k\pi}{4} < \pi[/tex]
solves k < 2.8 so if where an integer k can take the ranges 1 to 2.

BUT if we put 2 in:
[tex]\frac{1.1902 + 4\pi}{4}[/tex] this is greater than [tex]\pi[/tex]

Thanks
Thomas
 
Physics news on Phys.org
  • #2
thomas49th said:
To lie in the second quadrant [tex]\theta < \pi[/tex]

[tex]\frac{1.1902 + 2k\pi}{4} < \pi[/tex]
solves k < 2.8

No it doesn't, try solving this inequality again, and show your steps if you get the same result.
 
  • #3
im an idiot :blushing:
 

What is the Argand diagram?

The Argand diagram, also known as the Argand plane, is a graphical representation of complex numbers. It is named after Jean-Robert Argand, who first introduced the diagram in 1806. The Argand diagram is a two-dimensional Cartesian coordinate system where the real part of a complex number is represented on the x-axis and the imaginary part is represented on the y-axis.

What is De Moivre's theorem?

De Moivre's theorem, also known as the de Moivre-Laplace theorem, is a mathematical theorem that relates complex numbers and trigonometry. It states that for any positive integer n and any complex number z, (cos θ + i sin θ)n = cos nθ + i sin nθ, where θ is the angle formed by the complex number z and the positive real axis on the Argand diagram. This theorem is useful for simplifying the computation of powers of complex numbers.

How do you use De Moivre's theorem to find the nth root of a complex number?

To find the nth root of a complex number using De Moivre's theorem, first convert the complex number to polar form (r(cos θ + i sin θ)). Then, use De Moivre's theorem to raise the modulus r to the power of 1/n and divide the angle θ by n. The result will be the nth root of the complex number in polar form. To convert it back to rectangular form, use the trigonometric identities cos θ = (e^(iθ) + e^(-iθ))/2 and sin θ = (e^(iθ) - e^(-iθ))/2i.

What are the applications of De Moivre's theorem?

De Moivre's theorem is used in various fields of mathematics, including complex analysis, trigonometry, and algebra. It is also used in physics and engineering to solve problems involving periodic functions and differential equations. In addition, De Moivre's theorem has applications in computer graphics, signal processing, and cryptography.

Can De Moivre's theorem be extended to non-integer powers?

Yes, De Moivre's theorem can be extended to non-integer powers using the concept of complex exponentiation. For a complex number z and a real number α, the expression z^α is defined as e^(αln z), where ln z is the natural logarithm of z. This allows us to use De Moivre's theorem for any value of α, including fractions and irrational numbers.

Suggested for: Getting the argand with De Moires theorem

Replies
3
Views
325
Replies
3
Views
517
Replies
4
Views
721
Replies
26
Views
1K
Replies
2
Views
1K
Replies
23
Views
1K
Replies
5
Views
992
Replies
7
Views
800
Back
Top