I'll do it the algebra way, and then I'll do it the trigonometric way.

The ALGEBRA way:

4·sqrt(3+4i) = A + Bi,  where A and B are real.

16(3 + 4i) = A² + 2ABi + B²i²

48 + 64i = A² + B²i² + 2ABi

48 + 64i = A² + B²(-1) + 2ABi

48 + 64i = A² - B² + 2ABi

48 = A² - B²
64 = 2AB

Solve the second equation for B:  B =  = 

Substitute in 48 = A² - B²

48 = A² - ()²

48 = A² - )

48A² = A4 - 1024

0 = A4 - 48A - 1024

0 = (A² - 64)(A² + 16}}}

A² - 64 = 0;   A² + 16 = 0
     A² = 64        A² = -16
      A = ±8         A = ±4i (discard since A is real)

Using A = -8, = B =  =  = -4

So one solution is A + Bi = -8 - 4i

Using A = 8, = B =  =  = 4

So the other solution is A + Bi = 8 + 4i

-------------------
 
The trigonometric way: 
   ______
±4√3 + 4i
                  ______
We'll just get  ±√3 + 4i and multiply by 4 when we finish:

Get 3 + 4i in trigonometric form:
     _______    ______    __
r = V3² + 4² = V9 + 16 = V25 = 5
θ = arctan( = 53.13° + k·360°

Trig form is r(cos(θ+k*"350°") + isin(θ+k*"360°")), 

3 + 4i = 5(cos(53.13°+k·360°) + i·sin(53.13°+k·360))

 = 

 =  =

 =

, where k = 0, 1.

Answer is 4 times that or:

, where k = 0, 1.

That works out to be:

When k = 0

, 

4(2.000000893 + .999992136i) or

8.000003573+3.999992854i

which is very close to 8+4i.  The difference is because we
rounded off the angle.

When k = 1

, 

4(-2.000000893 - .999992136i) or

-8.000003573-3.999992854i

which is very close to -8-4i.

Edwin