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18 votes
18 votes
Help me: Solve: z^4 = -4

All answers should be in the format a+bi where a and b are real numbers

User Zebedee
by
2.9k points

2 Answers

21 votes
21 votes

Answer:

z = {-1 -i, -1 +i, 1 -i, 1 +i}

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.

User CESCO
by
3.3k points
9 votes
9 votes

z²-2i=0

Answer:

Solution given:

z⁴=-4

doing square root on both side


√(z⁴)=√(2²*-1)

we know i²=-1

substituting i² we get

z²=
√(2²*i²)

z²=2i

z²-2i=0 is a required answer. where

z² and 2 is real number.

User Roysch
by
3.0k points