Question 1043673
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Please help me solve (12i)^1/2 to a+bi form
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The general procedure on how to find the roots of a complex number is explained in the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/How-to-take-a-root-of-a-complex-number.lesson>How to take a root of a complex number</A>

in this site.


If you just are familiar with complex numbers, operations on them, complex plane, trigonometric form of complex numbers - 
&nbsp;&nbsp;&nbsp;&nbsp;- then you will be able to understand it.


If you are not familiar with this material, then you can learn on complex numbers from these lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Complex-numbers-and-arithmetical-operations.lesson>Complex numbers and arithmetic operations on them</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Complex-plane.lesson>Complex plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Addition-and-subtraction-of-complex-numbers-in-complex-plane.lesson>Addition and subtraction of complex numbers in complex plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Multiplication-and-division-of-complex-numbers-in-complex-plane-.lesson>Multiplication and division of complex numbers in complex plane</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/Raising-a-complex-number-to-an-integer-power.lesson>Raising a complex number to an integer power</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/complex/How-to-take-a-root-of-a-complex-number.lesson>How to take a root of a complex number</A>


<pre>
After this introduction, let me briefly explain you how to solve your problem.

So, you need to find {{{sqrt(12*i)}}}.

Write 12*i in the trigonometric form z = {{{r*(cos(alpha) + i*sin(alpha))}}},

where "r" is the modulus and {{{alpha}}} is the argument (polar angle).

In your case,  z = 12*i  in trigonometric form is z = {{{12*(cos(pi/2) + i*sin(pi/2))}}}, so the modulus is r = 12 and the polar angle is {{{alpha}}} = {{{pi/2}}}.

Now, to find the square root of this complex number, you have

  1.  to take a square root of the modulus: {{{sqrt(r)}}} = {{{sqrt(12)}}} = {{{2*sqrt(3)}}}.

  2.  to divide the argument (polar angle) by 2:  {{{alpha/2}}} = {{{((pi/2))/2}}} = {{{pi/4}}}.

  3.  to consider the complex number {{{z[1]}}} = {{{sqrt(r)*(cos(alpha/2) + i*sin(alpha/2))}}}, which is in your case 

      {{{z[1]}}} = {{{sqrt(12)*(cos(pi/4) + i*sin(pi/4))}}} = {{{2*sqrt(3)*(sqrt(2)/2 + i*sqrt(2)/2)}}} = {{{sqrt(3)*(sqrt(2) + i*sqrt(2))}}} = {{{sqrt(6)*(1 + i))}}} = {{{sqrt(6) + i*sqrt(6))}}}.

      It is one of the two complex roots.  // Notice that the modulus of {{{z[1]}}} is {{{sqrt(r)}}} and the argument is {{{alpha/2}}} = {{{pi/4}}}.

                                          // Also notice that the final expression for {{{z[1]}}} is just a + bi form.

  4.  to get the second root {{{z[2]}}} in trigonometric form, you have to use the same modulus as {{{z[1]}}} has, namely {{{sqrt(r)}}},  but use another 
      argument, which this time is {{{alpha/2 + 2pi/2}}} = {{{alpha/2 + pi}}}.

      Then your {{{z[2]}}} = {{{sqrt(r)*(cos(alpha/2 + pi) + i*sin(alpha/2 + pi))}}} = {{{sqrt(12)*(cos(pi/4 + pi) + i*sin(pi/4 + pi))}}} = {{{2*sqrt(3)*(cos(5pi/4) + i*sin(5pi/4))}}} = 

                     = {{{2*sqrt(3)*((-sqrt(2)/2) + i*(-sqrt(2)/2))}}} = {{{sqrt(3)*(-sqrt(2) - i*sqrt(2))}}} = {{{-sqrt(3)*(sqrt(2) + i*sqrt(2))}}} = {{{-sqrt(6)*(1 + i)}}} = {{{-sqrt(6) - i*sqrt(6)}}}.

      // Notice that {{{z[2]}}} = {{{-z[1]}}}.
      // All this long way with {{{z[2]}}} lead us to the opposite number to {{{z[1]}}}.
      // But now you know all the procedure, how it works for square roots of complex numbers.
      // Surely, it may seem too complex, at the first glance.
      // But there is a powerful symmetry in it, which work nicely for all n > 2.
      // If you read the lessons I recommended you, you will be able to learn its real power and beauty.


<U>Answer</U>.  {{{sqrt(12*i)}}} has two values: {{{z[1]}}} = {{{sqrt(6) + i*sqrt(6))}}} and {{{z[2]}}} = {{{-z[1]}}} = {{{-sqrt(6) - i*sqrt(6))}}}.
</pre>