Square Root of a complex number

Question

sqr root (-15+8i)

Answer 1

The trigonometric form of -15+8i is, (8,15,17 is a right triangle)

-15+8i=17(cosθ+isinθ), where θ=arctan(-8/15).

The angle is in the second quadrant and cosθ=-15/17, sinθ=8/17. 

For the square root we have

√(-15+8i)=√17(cosθ/2+isinθ/2)

cosθ/2=√((1+cosθ)/2)=√((2/17)/2)=1/√17

sinθ/2=√((1-cosθ)/2)=√(32/17)/2=4/√17

√(-15+8i)=±√17(1/√17+i4/√17)=±(1+4i)

sorry for leaving out one square root

This comes from z=re^iθ+2kπ,z^1/2=r^1/2e^iθ/2+kπ. k=1 is the only value to produce a different answer, the one with the minus sign.

(±(1+4i))²=1-16+8i=-15+8i AT LAST!

Apologies for my premature attempt at correction.

Answer 2

This is a trigonometry problem in which you convert the complex number into its Euler number being the form of:

 

A*e^ix = A*(cos(x)+i*sin(x)) where A is the magnitude of the vector in the complex plane in which is:

 

A = sqrt((-15)²+8²) = sqrt(225+64) = sqrt(289)

 

Factor out sqrt(289) from the original complex number in which is:

 

sqrt(289)*(-15/sqrt(289)+8i/sqrt(289))

 

Determine, the angle in in which the vector makes in the complex plane in which is:

 

x = tan^-1(8/-15) Note: Ask yourself what quadrant the vector lies in determining the proper angle.

 

The square root of a complex number simply becomes the square root of  the Euler number in which is:

 

sqrt(A*e^ix) = ±e^±ix/2*sqrt(A) = ±sqrt(A)*(cos(x/2) ± i*sin(x/2))

 

 

I'll let you finish this problem.