Respuesta :

msm555

z²-2i=0

Answer:

Solution given:

z⁴=-4

doing square root on both side

[tex]\sqrt{z⁴}=\sqrt{2²*-1}[/tex]

we know i²=-1

substituting i² we get

z²=[tex]\sqrt{2²*i²}[/tex]

z²=2i

z²-2i=0 is a required answer. where

z² and 2 is real number.

sqdancefan

Answer:

  z = {-1 -i, -1 +i, 1 -i, 1 +i}

Step-by-step explanation:

Using Euler's formula, we can write the equation as ...

  z^4 = 4e^(i(π +2πn)) . . . . . . where n is any integer

Then the 4th root is ...

  z = (√2)e^(i(π +2πn)/4 = (√2)e^(i(π/4 +nπ/2)) . . . . . for n = 0–3

  z = √2(cos(π/4+nπ/2) +i·sin(π/4 +nπ/2)) . . . . . for n = 0–3

  z = ±1 ±i

  z = {1 +i, -1 +i, -1 -i, 1 -i}

_____

Additional comment

Values of n are only needed in the range 0–3, because the answers repeat for values other than those.

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