SOLUTION: Find the cube roots of the following complex number; -5(square root of 2)+ 5(square root of 2i)

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Question 1073428: Find the cube roots of the following complex number;
-5(square root of 2)+ 5(square root of 2i)

Found 2 solutions by ikleyn, KMST:
Answer by ikleyn(52805)   (Show Source): You can put this solution on YOUR website!
.
Check if you correctly wrote parentheses.


Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
I believe what you wanted was the cube roots of
.
You can write that as
-5sqrt(2)+i5sqrt(2) or as -5*sqrt(2)+i*5*sqrt(2) .


The three cube roots will be numbers of the form
such that
,
meaning that for
.
So, gives us
, and .
You can get approximate values for
and for the trigonometric functions from a calculator.




Finding exact values is a little more involved,
and the resulting expressions may not look as nice as with approximate decimals.
The one answer with is easy,
because ,
so that the exact value for one of the three cube roots is
.

For the other two values of , it gets a little more complicated.
Since and ,
we can use the trigonometric functions for ,
knowing their relation to the trigonometric functions for and .
For second quadrant ,
and .
For fourth quadrant ,
and .

If we want exact values, we will need to use the half-angle trigonometric formulas:
and .
, so
and .
So, and
That makes the exact values for the other two cube roots
= and
= .

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