+0  
 
0
42
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avatar+133 

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!

 May 19, 2020
 #1
avatar+23566 
+1

z^2 = 2i

z = +- sqrt 2i =   1+i      or  1 -i    or   -1 -i     or   -1 + i

 May 19, 2020
 #2
avatar+133 
-1

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
avatar+20883 
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  180o  (if you wish, you can change this into radians).

 

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

 

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  41/4cis( 180o/4 )  =  sqrt(2)·cis( 45o )

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

of the primary fourt root three times:

So, your four fourth roots are:

sqrt(2)·cis( 45o )

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

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

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

 

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

sqrt(2)·cis( 135o )  =  -1 + i

sqrt(2)·cis( 225o )  =  -1 - i

sqrt(2)·cis( 315o )  =  1 - i

 May 19, 2020

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