Let Z be a complex number.whicj of the following is solution set of z^3-iz=0
2 answers
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Don't see any "following", but expect to see something like ....
z^3-iz=0
z(z^2 - i) = 0
z = 0 or z = ±√ i
Now about that √ i
let a + bi = √ i
a^2 + 2abi + b^2 i^2 = i
a^2 - b^2 + 2abi = 0 + i
a^2 - b^2 = 0
a = ±b
and
2abi = i
2ab = 1
2a^2 = 1
a = ±1/√2 = ±√2 / 2
z=0 , z = ±√2 / 2 ± (√2 / 2) i = ±(√2/2)(1 + i)
z^3-iz=0
z(z^2 - i) = 0
z = 0 or z = ±√ i
Now about that √ i
let a + bi = √ i
a^2 + 2abi + b^2 i^2 = i
a^2 - b^2 + 2abi = 0 + i
a^2 - b^2 = 0
a = ±b
and
2abi = i
2ab = 1
2a^2 = 1
a = ±1/√2 = ±√2 / 2
z=0 , z = ±√2 / 2 ± (√2 / 2) i = ±(√2/2)(1 + i)