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

Note: All solutions should be expressed in the form a+bi, where a and b are real numbers.Please write your answer in complete sentences.

Thanks in Advance!

CPhilFanboy May 19, 2020

#1

#2**-2 **

Thanks Pavlov! If it's possible could you edit and write it in sentences so I could break it down and learn from it.Thanks!

CPhilFanboy
May 19, 2020

#3**0 **

EP is off-line; I'll try to explain it in a way that makes sense to me.

To find the four fourth roots of -4: z = (-4)^{1/4}

I'm assuming that you are familiar with r·cis( theta ) form; if not, this may not make much sense.

If you graph z = -4 on the complex plane, you'll notic that its distance from the origin is 4;

this means that r = 4

You'll also notice that it's directly west; therefore, its angle is 180^{o} (if you wish, you can change this into radians).

So, the complex number -4 = -4 + 0i = 4cis(180^{o})

To find the primary fourth root, find the fourth root of r and divide the angle by 4.

So, the primary fourth root of -4 is 4^{1/4}cis( 180^{o}/4 ) = sqrt(2)·cis( 45^{o} )

To find the other fourth roots, divide 360^{o} by 4 to get 90^{o} and add this amount onto the angle

of the primary fourt root three times:

So, your four fourth roots are:

sqrt(2)·cis( 45^{o} )

sqrt(2)·cis( 45^{o} + 90^{o} ) = sqrt(2)·cis( 135^{o} )

sqrt(2)·cis( 135^{o} + 90^{o} ) = sqrt(2)·cis( 225^{o} )

sqrt(2)·cis( 225^{o} + 90^{o} ) = sqrt(2)·cis( 315^{o} )

sqrt(2)·cis( 45^{o} ) = sqrt(2)·cos( 45^{o} ) + i·sqrt(2)·sin( 45^{o} ) = 1 + 1i

sqrt(2)·cis( 135^{o} ) = -1 + i

sqrt(2)·cis( 225^{o} ) = -1 - i

sqrt(2)·cis( 315^{o} ) = 1 - i

geno3141 May 19, 2020