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Find all complex numbers z such that z^2 = 2i. Write your solutions in a+bi form, separated by commas. smiley

 Aug 4, 2022
 #1
avatar+124596 
+2

{z = a + bi }

 

 

( a + bi)^2  = 2i

 

a^2  + 2abi + (bi)^2  =  0 +  2i

 

a^2 - b^2  + 2abi  =   0 + 2i

 

Equating terms   .....This implies that

 

ab =  1

 

a = 1/b        .... so ....

 

a ^2 - b^2  =  0

 

a^2  - (1/a)^2  =  0

 

a^2  =  1/a^2                   multiply both sides by a^2

 

a^4  = 1      

 

a =  1                                 or      a  =  - 1

and b = 1/a = 1                    and b = 1/a  = -1                    

 

So

 

We have two  solutions

 

a + bi  =     1 + i

 

and

 

a + bi  =  -1 - i

 

Hope this makes  sense !!

 

cool cool cool

 Aug 4, 2022
edited by CPhill  Aug 4, 2022
 #2
avatar+177 
+6

Yes, this makes much more sense, thanks! smiley smiley smiley

waterlili  Aug 4, 2022

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