Calculating and Graphing the 4th Root of -4

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To find the 4th root of -4, it is essential to express -4 in complex exponential form, where z = -4 can be represented as r e^{i θ} with r = 4 and θ = π. The calculation involves using Euler's formula to derive the nth roots, specifically z^{1/n} = r^{1/n} e^{i kθ / n}, where k ranges from 0 to n-1. A common mistake is miscalculating the argument of -4, which should be π, not 0. Understanding these concepts is crucial for accurately graphing and calculating the roots in the complex plane.
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Hello everyone.

How to find the 4th root of -4? I know it's just plugging in the number into the formula but how since n=4, how can we calculate that without calculator? And how to draw it? Here I attached what I have done so far.
 

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How you draw z=-4? What is the angle of -4? It is not 0 as you wrote.

ehild
 
It would help to express -4 in complex exponential form.
 
MissP.25_5 said:
Hello everyone.

How to find the 4th root of -4? I know it's just plugging in the number into the formula but how since n=4, how can we calculate that without calculator? And how to draw it? Here I attached what I have done so far.

You've made a simple mistake in your calculation of arg(-4).

z = -4 + i0, or (-4, 0)

θ = arctan (0/-4) = π

Even though z = -4, draw it on the complex plane properly.
 
SteamKing said:
You've made a simple mistake in your calculation of arg(-4).

z = -4 + i0, or (-4, 0)

θ = arctan (0/-4) = π

Even though z = -4, draw it on the complex plane properly.

Oh yes, that was a careless mistake. Ok, so now I have the values:
r=4
n=4
θ=∏

But how do I find 4√4 ?
 
You know that x^4= (x^2)^2, right? So \sqrt[4]{4}= \sqrt{\sqrt{4}}. What is the square root of 4? What is the square root of that?
 
SteamKing said:
Use this information and apply Euler's formula.

z = r e^{i θ}

z^{1/n} = r^{1/n} e^{i kθ / n}, 0\leq k \lt n

See:

http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Roots.aspx

If I compute z = r e^{i θ}, wouldn't that bring us back to the start? Because that is -4.
Could you elaborate please? I don't really get it.
 
HallsofIvy said:
You know that x^4= (x^2)^2, right? So \sqrt[4]{4}= \sqrt{\sqrt{4}}. What is the square root of 4? What is the square root of that?

Thanks! I never thought of that. But what if we were to find the 5th root? I don't think this method can be applied.
 
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MissP.25_5 said:
If I compute z = r e^{i θ}, wouldn't that bring us back to the start? Because that is -4.
Could you elaborate please? I don't really get it.

That's your problem in a nutshell. You are working with complex numbers and you don't understand what is going on.

SteamKing said:
Use this information and apply Euler's formula.

z = r e^{i θ}

z^{1/n} = r^{1/n} e^{i kθ / n}, 0\leq k \lt n

I wrote these two formulas as a reminder of

1. how to express any number in exponential form, using Euler's formula, and

2. how to find the n nth roots of said number.

Your original problem was to find the 4 fourth roots of -4, or in other words, solve the equation

z^{4}+4 = 0 or

z^{4}= -4

Let's say the solutions to this equation are the complex numbers

ω_{1}, ω_{2}, ω_{3}, and ω_{4}

By writing -4 in the form z = r e^{i θ}, where z = -4,
we can use the second formula from the quote,

z^{1/n} = r^{1/n} e^{i kθ / n}

to calculate the numerical values of ω

I really recommend that you study the article linked below very carefully:

http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Roots.aspx

For a better visual representation of the cyclic nature of such roots:

http://mathworld.wolfram.com/RootofUnity.html
 

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