Simplifying Complex Calculation

[jboyd536]

Calculate

( minus ( 2 over 3 ) + ( 2 over 3 ) i ) to the power minus 4,

simplifing your answer and giving it in the form a + i b, with a and b given exactly.I found the modulus by:
sqrt((-2/3)^2 + (2/3)^2)
= (2*sqrt(2))/3

the argument is:
pi - 1 (from a sketch in the complex plane)

hence:
-2/3 + 2/3i = (2*sqrt(2)/3)*(cos(pi-1)+isin(pi-1))

using de moivres formula:
(-2/3 + 2/3i)^4 = (2*sqrt(2)/3)^4*(cos(4(pi-1))+isin(4(pi-1)))

but what next? I know I need to convert to cartesian form but how?

 

[Ray Vickson]

Calculate

( minus ( 2 over 3 ) + ( 2 over 3 ) i ) to the power minus 4,

simplifing your answer and giving it in the form a + i b, with a and b given exactly.I found the modulus by:
sqrt((-2/3)^2 + (2/3)^2)
= (2*sqrt(2))/3

the argument is:
pi - 1 (from a sketch in the complex plane)

hence:
-2/3 + 2/3i = (2*sqrt(2)/3)*(cos(pi-1)+isin(pi-1))

using de moivres formula:
(-2/3 + 2/3i)^4 = (2*sqrt(2)/3)^4*(cos(4(pi-1))+isin(4(pi-1)))

but what next? I know I need to convert to cartesian form but how?

How do you write $r e^{i \theta} = r \cos(\theta) + i r \sin(\theta)$ in the form $x + iy$?

 

[SteamKing]

Calculate

( minus ( 2 over 3 ) + ( 2 over 3 ) i ) to the power minus 4,

simplifing your answer and giving it in the form a + i b, with a and b given exactly.I found the modulus by:
sqrt((-2/3)^2 + (2/3)^2)
= (2*sqrt(2))/3

This looks OK.

the argument is:
pi - 1 (from a sketch in the complex plane)

How did you determine this? Show your sketch, if necessary.

hence:
-2/3 + 2/3i = (2*sqrt(2)/3)*(cos(pi-1)+isin(pi-1))

using de moivres formula:
(-2/3 + 2/3i)^4 = (2*sqrt(2)/3)^4*(cos(4(pi-1))+isin(4(pi-1)))

but what next? I know I need to convert to cartesian form but how?

Here is a nifty graphic: