# Help please with De Moivre's Theorem (raising a complex number to a power)

• Eng studies
In summary: I don't know what this is relevant to. I don't see any '0' or other Greek letters in the problem statement.I hope you know that 'cos0' should be a theta, not a zero. Oh, you're right. I didn't see that. Thanks for pointing that out.Bear with me guys i could not get the symbols to loadIf you're having trouble with LaTeX, there is a "LaTeX and MathType Help" forum you can post in for help.
Eng studies
Homework Statement
Convert to polar form and find z'8 using de moivre theorem, plot on argand diagram
Relevant Equations
Demoivre - (cos0+jsin0)'n = cosn0 +jsinno
Hi all any help on this would be great I cant seem to progress with the theorem,

z= -2 + j > R sqrt (-2)'2 + (-1)'2
r = 2.24

0= Arctan(-1) = 26.57 Polar form = 2.24(cos(26.58)+jsin(26.58)
-2
Demoivre - (cos0+jsin0)'n = cosn0 +jsinno

Could some one work a similar example to assist me through applying this ?

Thanks

Eng studies said:
Homework Statement: Convert to polar form and find z'8 using de moivre theorem, plot on argand diagram
Relevant Equations: Demoivre - (cos0+jsin0)'n = cosn0 +jsinno

Could some one work a similar example to assist me through applying this ?
Have you tried a Google search for videos on this subject? I did a quick search and found a number of videos that look like they would be helpful for you.

I have had a quick look maybe ill spend some more time just struggling to apply to question set above

Eng studies said:
Homework Statement: Convert to polar form and find z'8 using de moivre theorem, plot on argand diagram
Relevant Equations: Demoivre - (cos0+jsin0)'n = cosn0 +jsinno
I hope you know that 'cos0' should be a theta, ##\theta##, not a zero. It is true for any angle, ##\theta##.
Demoivre: ##(\cos \theta + j \sin \theta)^n = \cos n\theta + j \sin n \theta##.
Eng studies said:
Hi all any help on this would be great I cant seem to progress with the theorem,

z= -2 + j > R sqrt (-2)'2 + (-1)'2
r = 2.24
I don't know what this is relevant to. Exactly what is the statement of the problem?
Eng studies said:
0= Arctan(-1) = 26.57 Polar form = 2.24(cos(26.58)+jsin(26.58)
-2
None of that makes sense to me. I don't know what it is relevant to.

WWGD, PeroK and berkeman
FactChecker said:
I hope you know that 'cos0' should be a theta, not a zero.
Ohh, thanks for that. I could not decipher the OP's question to save my life.

Bear with me guys i could not get the symbols to load see attached, This is where im at if you could verify the polar conversion that would be great.

Following that I need some help applying this into the theorem any relevant assistance/workings would be appreciated

Workings attached with the correct symbols

to plot z8 I presume n needs to be power of 8 but I am getting really high numbers which i do not think are correct

#### Attachments

• Capture de moivre.PNG
6.6 KB · Views: 75
Eng studies said:
Bear with me guys i could not get the symbols to load
To post math equations, it's best to use the LaTeX engine that PF provides. There is a helpful "LaTeX Guide" link below the Edit window to get you started. Note that you put double-\$ delimiters at the start and end of each stand-alone line of LaTeX, and double-# delimiters at the start and end of in-line LaTeX that does not need to be on its own line. Also, if you right-click on a LaTeX equation in a post, you get a pop-up menu to let you view the LaTeX source or view it in other formats.

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Ok Perfect ill have a play around with that going forward, Does my question make sense to you now ? is the polar conversion correct ?

Eng studies said:
Homework Statement: Convert to polar form and find z'8 using de moivre theorem, plot on argand diagram
Relevant Equations: Demoivre - (cos0+jsin0)'n = cosn0 +jsinno

Hi all any help on this would be great I cant seem to progress with the theorem,
It appears that the problem statement is this:
If z = -2 + j, use the Theorem of De Moivre to calculate ##z^8## and plot this as an Argand diagram.
Eng studies said:
z= -2 + j > R sqrt (-2)'2 + (-1)'2
r = 2.24
Do not use ' to indicate an exponent. If you don't use LaTeX, the symbol ^ is often used for this purpose.
Eng studies said:
0= Arctan(-1) = 26.57 Polar form = 2.24(cos(26.58)+jsin(26.58)
-2
Demoivre - (cos0+jsin0)'n = cosn0 +jsinno
The above is the form for complex numbers with modulus equal to 1. As was already mentioned, don't use 0 in place of ##\theta## (theta).
##(\cos(\theta) + j\sin(\theta))^n = \cos(n\theta) + j(\sin(n\theta)##
There is a different version for complex numbers whose modulus is not equal to 1.
Eng studies said:
Could some one work a similar example to assist me through applying this ?

Thanks
Here's an example.
If z = 1 + j, find ##z^3##

Solution
##|z| = \sqrt{1^2 + 1^2} = \sqrt 2##
##\theta = \arctan(1/1) = \frac \pi 4##
In polar form ##z = \sqrt 2(\cos(\pi/4) + j\sin(\pi/4))##
Then ##z^3 = (\sqrt 2)^3(\cos(3\pi/4) + j\sin(3\pi/4))##

To plot ##z^3##, the argument (angle) is 135° (halfway into Quadrant II) and the magnitude is ##2\sqrt 2 \approx 2,828##.

Eng studies, jim mcnamara and berkeman
In your work shown in the attachment, you found an angle in Quadrant IV. That is incorrect- it should be in Quadrant II since that's where z is.

If you have a negative value [real or imaginary] the atan does not return the actual argument. In your case the argument is ACOS(-2/2.24) only.

If you have a negative value [real or imaginary] the atan does not return the actual argument.
In this case, the actual argument is ##\tan^{-1}(-2) + \pi## or in degrees, about 153.4°.
In your case the argument is ACOS(-2/2.24) only.
More precisely, ##\cos^{-1}(\frac{-2}{\sqrt 5})##.

Thanks all I will read through this in detail and have another attempt will post progress when done

find the angle, multiply it by 8, and then raise the length to the 8th power. that gives angle and length. then use trig to plot it in coords (i.e. find cos and sin of the angle, then scale by the length).

Last edited:
FactChecker
I ended up getting 625(cos154.48 + jsin154.48) but I am having a hard time understanding how 625 is plotted on an argand diagram?

ckfo said:
I ended up getting 625(cos154.48 + jsin154.48) but I am having a hard time understanding how 625 is plotted on an argand diagram?
To plot a complex number of the form ##A(\cos(\theta) + i\sin(\theta))##, rotate counter-clockwise through an angle of ##\theta## (measured from the positive real axis), and then go out that ray A units.

Gavran
ckfo said:
I ended up getting 625(cos154.48 + jsin154.48) but I am having a hard time understanding how 625 is plotted on an argand diagram?
That's just the length/modulus of the number. The length of the line joining the origin with the ( representation of the) number.

## What is De Moivre's Theorem?

De Moivre's Theorem states that for any complex number in polar form $$z = r(\cos \theta + i \sin \theta)$$ and any integer $$n$$, the power of the complex number can be expressed as $$z^n = r^n (\cos (n\theta) + i \sin (n\theta))$$. This theorem is useful for raising complex numbers to integer powers.

## How do you convert a complex number to polar form?

To convert a complex number $$z = a + bi$$ to polar form, you need to find the magnitude $$r$$ and the angle $$\theta$$. The magnitude is $$r = \sqrt{a^2 + b^2}$$ and the angle $$\theta$$ is $$\theta = \tan^{-1}(\frac{b}{a})$$. The polar form is then $$z = r(\cos \theta + i \sin \theta)$$.

## Can De Moivre's Theorem be used for non-integer powers?

De Moivre's Theorem is specifically formulated for integer powers. For non-integer powers, the theorem does not directly apply, and more advanced techniques involving complex logarithms and fractional exponents are required.

## How do you apply De Moivre's Theorem to find the power of a complex number?

First, express the complex number in polar form $$z = r(\cos \theta + i \sin \theta)$$. Then, apply De Moivre's Theorem by raising the magnitude to the power $$n$$ and multiplying the angle by $$n$$: $$z^n = r^n (\cos (n\theta) + i \sin (n\theta))$$. Finally, you can convert the result back to rectangular form if needed.

## What are some practical applications of De Moivre's Theorem?

De Moivre's Theorem is used in various fields such as signal processing, quantum mechanics, and electrical engineering. It simplifies the computation of powers and roots of complex numbers, which are essential in solving differential equations, analyzing waveforms, and studying oscillatory systems.

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