Respuesta :
Answer:
D. 64 (cos(pi)+isin(pi))
Step-by-step explanation:
Put 2+2i into trigonometric form, raise the r value to the power of 4, and multiply the angle value (theta) by 4 to get the equation of z^4. In this case, r was 2(sqrt2), which raised to the 4th power was 64. The angle value was pi/4, which multiplied by 4 was pi. That's how we get the answer:
64 (cos(pi)+isin(pi)). How this helps!
Following are the calculation to the given value:
Given:
[tex]\to Z= 2+2i[/tex]
To find:
[tex]Z^4=?[/tex]
Solution:
[tex]\to Z= 2+2i[/tex]
[tex]\therefore\\\\i^2= -1[/tex]
[tex](a+b)^2= a^2+b^2+2ab[/tex]
Square the given value:
[tex]\to Z^2= (2+2i)^2[/tex]
[tex]= 2^2+(2i)^2+2 \times 2 \times 2i\\\\= 4+4i^2+8i\\\\= 4-4+8i\\\\=8i[/tex]
When [tex]Z^2= 8i[/tex] then again square the value
[tex]\to ( Z^2)^2= (8i)^2\\\\\to Z^4= 64i^2\\\\\to Z^4= -64\\\\[/tex]
Therefore, the final answer is "-64".
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