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

1. Apr 11, 2010

### cortozld

1. The problem statement, all variables and given/known data
find the square root of: 4+4(sqrt3)i

put the answer in a+bi form

2. Relevant 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

3. 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

2. Apr 11, 2010

### Staff: Mentor

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.

3. Apr 13, 2010

### cortozld

Woo hoo! Thanks for the help

4. Apr 14, 2010

### HallsofIvy

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 $a^2- 12/a^2= 4$.

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