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Find all complex numbers z that satisfy z^2 = i/2.

 Jan 29, 2021
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Hi Guest!

 

z = a + bi

(a + bi)^2 = i/2

a^2 - b^2 - 2abi = i/2

2a^2 - 2b^2 - 4abi = i

2a^2 - 2b^2 = 0

a^2 = b^2 (Important)

4abi = i

ab = 1/4 (Important)

 

So |a| and |b| are the same number. 

And since ab = 1/4 (positive), they must be both negative or both positive. 

So a = 1/2 and b = 1/2 or a = -1/2 and b = -1/2. 

 

z = 1/2 + i/2

z = -1/2 - i/2

I hope this helped. :)))

 

=^._.^=

 Jan 29, 2021

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