Calculating and Graphing the 4th Root of -4

[MissP.25_5]

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.

 

[ehild]

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

ehild

 

[Fightfish]

It would help to express -4 in complex exponential form.

 

[SteamKing]

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.

 

[MissP.25_5]

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 ?

 

[HallsofIvy]

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]

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 ?

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

 

[MissP.25_5]

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.

 

[MissP.25_5]

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.

 

[SteamKing]

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.

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