+0

# math question

0
78
7

Find all complex numbers z such that z^4 = -4.

Dec 16, 2019

#1
+19741
0

z^4 = -4

z^2 = +- sqrt(-4) = +-2i

z = +- sqrt(+-2i)

=  -1+i  ,   -1-i    ,  1-i    ,  1+i

Dec 16, 2019
#2
+43
-1

For styling purposes:

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

whoisjoe  Dec 16, 2019
#5
+1

Question: is there a way to prove that those are ALL of the numbers? Thanks

Guest Dec 17, 2019
#6
+106885
+1

I personally do not know how to prove it but with all roots there is always 'that' number of roots,

so square root has 2 values

cube roots have 3 values

4th roots have 4 values  etc

I am sure if you google it you will find a proof.

Melody  Dec 17, 2019
#7
+1

Got it. Thanks Melody!

Guest Dec 18, 2019
#3
+1

Let's make the solution super-formal!  We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

Dec 17, 2019
#4
+1

that... is a more complicated way. BUT AWESOME!

Guest Dec 17, 2019