#1

#7**0 **

There actually are square roots for negative numbers- just not real numbers. Read the problem. It specifically says "complex numbers". Complex numbers have imaginary parts to it.

thelizzybeth
Jun 14, 2020

#9**+1 **

jimkey17:

There are square roots for negative numbers. Mathematicians defined the complex unit \(i\) such that \(i^2 = -1\), which in other words, \(i = \sqrt{-1}\). The set of all complex numbers is isomorphic to the set \(\left\{(a, b) \in\mathbb R^2: \begin{pmatrix}a&-b\\b&a\end{pmatrix}\right\}\).

Here is a related article on Wikipedia: https://en.wikipedia.org/wiki/Complex_number

MaxWong
Jun 14, 2020

#10**+1 **

Attn: theLizzybeth,

I have looked at your history and seen that you have provided many quality answers. I thank you for that.

Jimkey is also a valuable member of this forum.

I share Max's view that Jimkey has had no prior experience with complex numbers. Hence he has been taught that there are no square roots for negative numbers. Certainly, in the real number system, he is correct.

Please be aware that people are at different places in their learning.

Thanks Max ;)

Melody
Jun 15, 2020

#2**0 **

I'm assuming you mean \({z^2=-4}\)..!

Taking the square root if both sides we get \( {\pm(z)}=\pm\sqrt(-4)\).

so since \( {\sqrt(4)\sqrt(-1)=\sqrt(4)i=2i}\)(remember \({\sqrt(-1)=i}\))

we add the plus/minus sign back on so \({z=\pm(2i) }\)for all complex numbers

Guest Jun 14, 2020

#6**+1 **

Solve this as a polynomial, complete the square, and factorize.

First write it as \(z^4+4=0\)

Complete the square. \(z^4+4z^2+4-4z^2=(z^2+2)^2-(2z)^2=(z^2+2z+2)(z^2-2z+2)\)

We can factorize these two quadratics using \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) to find our answer (four complex numbers):

\(-1+i, -1-i, 1+i, 1-i\) or \(-1\pm i, 1\pm i\)

thelizzybeth Jun 14, 2020