Find the modulus and argument of sqrt(3+i).

Nov 2014
27
1
math
I tried letting sqrt(3+i) = a+bi. With some calculations, I got 2a^2 = x + sqrt(x^2+y^2), 2b^2 = -x + sqrt(x^2+y^2), which means a = +- sqrt[(3+sqrt(10))/2] and b = +- sqrt[(-3+sqrt(10))/2]. Does this mean there are four possible answers?

But then when I tried doing this by expressing the complex number in polar coordinate form, there are only two roots.

sqrt(3+i) = 10^(1/4) [cos (1/2)tan-1(1/3) + i sin (1/2)tan-1(1/3)]

or 10^(1/4) {cos [(1/2)tan-1(1/3)+pi] + i sin [(1/2)tan-1(1/3)+pi]}

What's wrong?
 

mathman

Forum Staff
May 2007
6,932
774
ab = 1/2, so the allowable solutions in the first approach are those where a and b have the same sign.
 
  • Like
Reactions: 1 person