# Cuckoo For Conjugates 2!

## The square root of a negative pure imaginary number is equal to the conjugate of the square root of its additive inverse!

## √(-b*i*) = the conjugate of √(b*i*)

__Examples:__

√(-*i*) = √(2)/2 - √(2)/2*i* & √(*i*) = √(2)/2 + √(2)/2*i*

√(-9*i*) = 3√(2)/2 - 3√(2)/2*i* & √(9*i*) = 3√(2)/2 + 3√(2)/2*i*

√(-5*i*) = √(5/2) - √(5/2)*i* & √(5*i*) = √(5/2) + √(5/2)*i*

√(-7*i*) = √(7/2) - √(7/2)*i* & √(7*i*) = √(7/2) + √(7/2)*i*

√(-1.5*i*) = √(3/2) - √(3/2)*i* & √(1.5*i*) = √(3/2) + √(3/2)*i*

√(-π*i*) = √(π/2) - √(π/2)*i* & √(π*i*) = √(π/2) + √(π/2)*i*

It doesn't matter what real number you plug in for the variable **b**, this trick will still work out! And remember: **real × imaginary = imaginary!**

According to the examples above, the real part & the imaginary part of each complex number is likely to be a fraction with the number **2** as the denominator! But if you pick an even number, then...

√(-2*i*) = 1 - *i* & √(2*i*) = 1 + *i*

√(-8*i*) = 2 - 2*i* & √(8*i*) = 2 + 2*i*

√(-36*i*) = 3√(2) - 3√(2)*i* & √(36*i*) = 3√(2) + 3√(2)*i*

...The real part & the imaginary part won't be fractions, but they're still likely to be multiples of *the square root of *2. Furthermore, both parts of the complex number will always equal each other.

## Finally, an extra fact to remember: If the real part & the imaginary part of a complex number are equal to each other, then the square of the complex number will be a pure imaginary number.

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© Derek Cumberbatch