# How do you find the square root of an imaginary number of the form a+bi?

##### 1 Answer

Alternatively, solve without using trigonometry to find that the square roots of

#+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)#

#### Explanation:

Suppose

How do we solve for

#(c+di)^2 = c^2+2cdi + d^2i^2 = (c^2-d^2) + (2cd)i#

So we want to solve:

#c^2-d^2 = a#

#2cd = b#

From the second of these, we find:

#d = b/(2c)#

So:

#d^2 = b^2/(4c^2)#

Substituting this in the first equation we get:

#c^2-b^2/(4c^2) = a#

Multiply through by

#4(c^2)^2-b^2 = 4ac^2#

Subtract

#4(c^2)^2-4a(c^2)-b^2 = 0#

From the quadratic formula, we find:

#c^2 = (4a+-sqrt((4a)^2+16b^2))/8 =(a+-sqrt(a^2+b^2))/2#

For

#c^2 = (a+sqrt(a^2+b^2))/2#

Hence:

#c = +-sqrt((sqrt(a^2+b^2)+a)/2)#

Then:

#d = +-sqrt(c^2-a)#

#= +-sqrt((sqrt(a^2+b^2)+a)/2-a)#

#= +-sqrt((sqrt(a^2+b^2)-a)/2)#

The remaining question is: What signs do we need to choose?

Since

If

If

If

If

#+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)#

**Footnote**

The question asked what is *the* square root of

For positive Real numbers *principal* square root of

For negative Real numbers *principal* square root is the one with a positive coefficient of

It becomes more complicated when we deal with the square roots of Complex numbers in general.

Consider:

#sqrt(-1/2-sqrt(3)/2i) = +-(1/2-sqrt(3)/2 i)#

Which sign do you prefer?