Asked by Trish Goal
Find all complex numbers such that
z^2=2i. Thank you!
z^2=2i. Thank you!
Answers
Answered by
Steve
it's easiest to use the polar form
if z = r cisθ
z^2 = r^2 cis2θ
so,
r^2 cis2θ = 2 cis π/2
r = ±√2
θ = π/4
So, we have
√2 cis π/4
-√2 cis π/4
√2/√2 + √2/√2 i = 1+i
or -(1+i)
check:
(1+i)^2 = 1+2i+i^2 = 1+2i-1 = 2i
same for -(1+i)
if z = r cisθ
z^2 = r^2 cis2θ
so,
r^2 cis2θ = 2 cis π/2
r = ±√2
θ = π/4
So, we have
√2 cis π/4
-√2 cis π/4
√2/√2 + √2/√2 i = 1+i
or -(1+i)
check:
(1+i)^2 = 1+2i+i^2 = 1+2i-1 = 2i
same for -(1+i)
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