Give a corect answer for this question how do I get the square root of a complex number? If given a complex number (say 9+4i), how do I calculate its square root?

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usaho4w · Accepted Answer

The square root is not a well defined function on complex numbers. Because of the fundamental theorem of algebra, you will always have two different square roots for a given number. If you want to find out the possible values, the easiest way is probably to go with De Moivres

Timothy Wolff · Accepted Answer

Here is a direct algebraic answer.  Suppose that z=c+di, and we want to find z=a+bi lying in the first two quadrants.  So what are a and b?  Precisely we have  a=c+c2+d22  and  b=d|d|−c+c2+d22  (The factor of d|d| is used so that b has the same sign as d) To find this, we can use brute force and the quadratic formula.  Squaring, we would need to solve   a2−b2+2abi=c+di.  This gives two equations and two unknowns (separate into real and  imaginary parts), which can then be solved by substitutions and the  quadratic formula.

nick1337 · Accepted Answer

You can also do following (technique often advised at school) : Let&#039;s write z2=9+4i with z=a+bi. The goal is to find z Thus we have (a+bi)2=9+4i and if you expend we get a2+2abi−b2=9+4i If you identify the real and imaginary parts, you obtain : a2−b2=9 (1) and 2ab=4 (2) Now, as z2=9+4i, the modulus of z2 and 9+4i are equal so we can write : a2+b2=92+42 a2+b2=97 (3) Now find a and b with the the equations (1) , (2) and (3) : K(1)+(3)⇒2a2=9+97 so a=12(9+97) or a=−12(9+9) With equation (2) and the previous result you can now find b : 2ab=4 b=2/a so b=229+97orb=−229+97 The answer is : z=12(9+97)+2i29+97orz=−12(9+97)−2i29+97

Give a corect answer for this question how do I

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