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

In summary: You can check your work by taking either of your square roots and squaring it. You should get 4 + 4√3i.Since it only asked for the square root you could also try(a+ bi)^2= a^2- b^2+ (2ab)i= 4+ 4i\sqrt{3}so that you must have a^2- b^2= 4 and 2ab= 4\sqrt{3}From 2ab= 4\sqrt{3}, b= 2\sqrt{3}/a and the first equation becomes a^2- 12/a^2= 4. Multip
  • #1
cortozld
11
0

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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  • #2
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
 
  • #3
Woo hoo! Thanks for the help
 
  • #4
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.
 

What is the square root of 4+4(sqrt3)i?

The square root of 4+4(sqrt3)i is 2+2(sqrt3)i.

How do you find the square root of a complex number?

To find the square root of a complex number, first write the number in the form a+bi, where a and b are real numbers. Then, use the formula sqrt(a+bi) = sqrt((sqrt(a^2+b^2)+a)/2) + sign(b)sqrt((sqrt(a^2+b^2)-a)/2)i. In this case, a=4 and b=4(sqrt3).

What is the real part of the square root of 4+4(sqrt3)i?

The real part of the square root of 4+4(sqrt3)i is 2.

What is the imaginary part of the square root of 4+4(sqrt3)i?

The imaginary part of the square root of 4+4(sqrt3)i is 2(sqrt3).

Is the square root of 4+4(sqrt3)i a real number or a complex number?

The square root of 4+4(sqrt3)i is a complex number.

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