Roots of complex number

Determine the roots in the following case :
 i. Square roots of i
ii. Square roots of (1+i)
iii. Square roots of 1+sqrt(3) i

Total no hints on how to solve the problem. ANy idea ?
Simon LeungAsked:
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d-glitchCommented:
There are two techniques:

1. Assume the form of the answer, and equate the coefficients.  
    Every complex number has the form z = a + bi, so that
    number squared will be   z² = (a²-b²)  + (2ab)i.

    For the first case: (a²-b²) = 0   and   (2ab) = 1
    Solve the two simultaneous equations for a and b.

2. Things get easier if you know about the polar representation
     of complex numbers.
phoffric\Commented:
Polar representation definitely makes it easier as D-glitch says.

i = e^[i*pi/2 +i*2pi*k] for k = any integer.

Then i^[1/2] =
e^[i*pi/4 +i*pi*k] for k = any integer.
phoffric\Commented:
e^[¡*x] = cos x + i*sin x
https://en.m.wikipedia.org/wiki/Euler%27s_identity#Imaginary_exponents

Letting k = 0 (although you can pick any integer value, but k = 0 is often chosen to keep the angle between 0 and 360 degrees):
i^[1/2] = e^[i*pi/4]
= cos pi/4 + i* sin pi/4

pi is 180 degrees, so pi/4 is 45 degrees.

cos 45 = sin 45 = 1/sqrt(2)

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Simon LeungAuthor Commented:
Can you comment whether my work (especially on last question) is correct or not ?

Thx.
MyWork.jpg
d-glitchCommented:
I think iii, iv, and v are wrong.

You have to calculate the magnitude and angle of the input.

Mag of   (1+i)  is  sqrt(2)

Mag of   (1+sqrt(3)i)  is  2
Simon LeungAuthor Commented:
any hint and more explanation on (v) Fourth roots of 1-i  ?

Another, rework iii & iv... any comment ?

Thx again.
c3a4b370-28e4-40ac-be26-fc2a9143245b.jpg
d-glitchCommented:
Still wrong.  You are getting the angles right, but not the magnitudes.

Mag of  (1+i)  is  sqrt(2)    Mag of answer will be sqrt( sqrt(2))  or 2^(1/4)


Mag of  (1-i)  is  sqrt(2)    Mag of answer will be 4thrt( sqrt(2))  or 2^(1/8)

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