Find the square root of: 4+4(sqrt3)i

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Homework Help Overview

The problem involves finding the square root of the complex number 4 + 4(sqrt3)i and expressing the answer in the form a + bi. The subject area pertains to complex numbers and their roots.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss converting the complex number to trigonometric form and applying the nth root formula. There are attempts to express the results in Cartesian form, with some questioning the accuracy of their conversions. One participant suggests an alternative approach using algebraic identities.

Discussion Status

Some participants have provided guidance on the conversion process and identified potential errors in the Cartesian form. There is an ongoing exploration of different methods to approach the problem, with no explicit consensus on a single solution path.

Contextual Notes

Participants express uncertainty about their calculations and seek further clarification on their methods. The discussion reflects a collaborative effort to understand the problem without providing a definitive solution.

cortozld
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Homework Statement


find the square root of: 4+4(sqrt3)i

put the answer in a+bi form

Homework Equations


finding the nth roots:

√(r(cos(theta)+isin(theta)))=√r(cos(theta+2(pi)(k)/(n))+isin(theta+2(pi)(k)/(n)

where k=0,1


The Attempt at a Solution



first i converted my equation to trigonometric form: 8(cos(pi/3)+isin(pi/3))

then using "finding the nth root" equation above i got: 2√2(cos(pi/6)+isin(pi/6) and 2√2(cos(7pi/6)+isin(7pi/6)

i then converted the two answers to a+bi form and got 2√6+√(2)i and -2√6-√(2)i

im pretty sure i did something wrong. If u need me to go more in depth on anything just post a comment
 
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cortozld said:

Homework Statement


find the square root of: 4+4(sqrt3)i

put the answer in a+bi form

Homework Equations


finding the nth roots:

√(r(cos(theta)+isin(theta)))=√r(cos(theta+2(pi)(k)/(n))+isin(theta+2(pi)(k)/(n)

where k=0,1


The Attempt at a Solution



first i converted my equation to trigonometric form: 8(cos(pi/3)+isin(pi/3))

then using "finding the nth root" equation above i got: 2√2(cos(pi/6)+isin(pi/6) and 2√2(cos(7pi/6)+isin(7pi/6)

i then converted the two answers to a+bi form and got 2√6+√(2)i and -2√6-√(2)i
Your answers in polar form look fine, but you made a mistake when you converted to Cartesian form.

2√2(cos(pi/6)+isin(pi/6)) = 2√2(√3 /2 + i/2) = √2(√3 + i) = √6 + i√2. The other root has a similar mistake.

You can check your work by taking either of your square roots and squaring it. You should get 4 + 4√3i.
cortozld said:
im pretty sure i did something wrong. If u need me to go more in depth on anything just post a comment
 
Woo hoo! Thanks for the help
 
Since it only asked for the square root you could also try
[itex](a+ bi)^2= a^2- b^2+ (2ab)i= 4+ 4i\sqrt{3}[/itex]
so that you must have [itex]a^2- b^2= 4[/itex] and [itex]2ab= 4\sqrt{3}[/itex]

From [itex]2ab= 4\sqrt{3}[/itex], [itex]b= 2\sqrt{3}/a[/itex] and the first equation becomes [itex]a^2- 12/a^2= 4[/itex].

Multiplying both sides by [itex]a^2[/itex], [itex]a^4-12= 4a^2[/itex] which is the same as [itex](a^2)^2- 4a^2- 12= 0[/tex]. Solve that quadratic equation for [itex]a^2[/itex] and then solve for a and b.[/itex]
 

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