(AM GIVING BRAINLIEST) Find all complex numbers z such that z^4 = -4 Note: All solutions should be expressed in the form a+bi, where a and b are real numbers. plz explain

  z = {1 +i, 1 -i, -1 +i, -1 -i}

Step-by-step explanation:

Using Euler's formula, you can write this as ...

  z^4 = 4e^(i(π+2kπ))

Then the 4th root is ...

  z = 4^(1/4)e^(i(π/4 +kπ/2)) = √2(cos(π/4+kπ/2) +i·sin(π/4+kπ/2))

for integers k in the range 0 to 3.

  z = √2(±1/√2 ±i·1/√2) . . . signs are independent

  z = ±1 ±i . . . . . . signs are independent; hence 4 solutions

  z = {1 +i, 1 -i, -1 +i, -1 -i}

Comment on roots of complex numbers

In the complex domain, each number has a number of roots equal to the index of the root. That is, there are four 4th roots. They all have the same magnitude, but their angles are separated by 360°/(root index) = 90°. Here, the principal fourth root of 4∠180° is 4^(1/4)∠(180°/4) = √2∠45° = 1+i.

Adding 90° is equivalent to multiplying by i, so the other 3 roots are √2∠135° = -1+i, √2∠225° = -1 -i, and √2∠315° = 1 -i.

 z = {1 +i, 1 -i, -1 +i, -1 -i}

can i see the following ?

step-by-step explanation: