SOLUTION: How do I find all the values to the following? 1) i^0.25 2) (1+sqrt(3)i)^(1/3) 3) (i-1)^0.5 4) ((9i)/(1+i))^(1/6)

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: How do I find all the values to the following? 1) i^0.25 2) (1+sqrt(3)i)^(1/3) 3) (i-1)^0.5 4) ((9i)/(1+i))^(1/6)      Log On


   



Question 1116154: How do I find all the values to the following?
1) i^0.25
2) (1+sqrt(3)i)^(1/3)
3) (i-1)^0.5
4) ((9i)/(1+i))^(1/6)

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The a%2Bi%2Ab form of complex numbers is not useful for calculating products, powers, and roots.
For those purposes the polar, or trigonometric, or exponential forms are much more useful.
For example, 1%2Bsqrt%283%29i can be written using
its absolute value or modulus, r=sqrt%281%5E2%2B%28sqrt%283%29%29%5E2%29=sqrt%281%2B3%29=sqrt%284%29=2 ,
and its argument theta such that tan%28theta%29=sqrt%283%29%2F1=sqrt%283%29 .
We could say theta=60%5Eo to keep it simple, or theta=pi%2F3 if you must use radians.
You can write 1%2Bsqrt%283%29i as 2%28cos%2860%5Eo%29%2Bi%2Asin%2860%5E0%29%29 or as e%5E%28i%2Api%28%221+%2F+3%22%29%29 .
All you need to remember is that when multiplying complex numbers,
absolute values multiply, and arguments add up.

2) So, the solutions to
are numbers of the form r%28cos%28theta%29%2Bi%2Asin%28theta%29%29 such that
.
That means that r%5E3=2<-->r=2%5E%221+%2F+3%22=root%283%2C2%29 ,
and 3theta is 60%5Eo or a coterminal angle.
So, the possibilities for theta between 0%5Eo and 360%5Eo are
3theta=60%5Eo --> theta=20%5Eo ,
3theta=60%5Eo%2B360%5Eo=420 --> theta=420%5Eo%2F3=140%5Eo ,
and 3theta=60%5Eo%2B720%5Eo=780 --> theta=780%5Eo%2F3=260%5Eo .
Approximate values are
,
so we could write the answers as
2%5E%221+%2F+3%22%28cos%2820%5Eo%29%2Bi%2Asin%2820%5Eo%29%29=approximately1.1839%2Bi%2A0.4309
2%5E%221+%2F+3%22%28cos%28140%5Eo%29%2Bi%2Asin%28140%5Eo%29%29=approximately-0.9652%2Bi%2A0.8099
2%5E%221+%2F+3%22%28cos%28260%5Eo%29%2Bi%2Asin%28260%5Eo%29%29=approximately0.2188-i%2A1.2408

1) The solutions to
are four fourth roots of 1%28cos%2890%5Eo%29%2Bi%2Asin%2890%5Eo%29%29 with
r=root%284%2C1%29=1 and 4theta=90%5Eo%2Bk%2A2pi%29 with k=%220+%2C+1+%2C+2+%2C+or+3%22 .
That means theta=22.5%5Eoor112.5%5Eoor202.5%5Eoor292.5%5Eo .
The approximate values of sine and cosine for those angles are:

So, the four complex values of i%5E0.25 are
cos%2822.5%5Eo%29%2Bi%2Asin%2822.5%29=approximately0.9239%2Bi%2A0.3827 ,
cos%28112.5%5Eo%29%2Bi%2Asin%28112.5%29=approximately-0.3827%2Bi%2A0.9239
cos%2822.5%5Eo%29%2Bi%2Asin%2822.5%29=approximately-0.9239-i%2A0.3827 , and
cos%2822.5%5Eo%29%2Bi%2Asin%2822.5%29=approximately0.3827-i%2A0.9239

3) %28i-1%29%5E0.5=sqrt%282%29%28cos%28315%5Eo%29%2Bi%2Asin%28315%5Eo%29%29%5E%221+%2F+2%22 has 2 values, with
r=sqrt%282%29%5E1%2F2=root%284%2C2%29=approximately1.1892 and 2theta=315%5Eo%2Bk%2A360%5Eo with k=%220+%2C+or+1%22 .
That means theta=%28315%5Eo%2Bk%2A360%5Eo%29%2F2=157.5%5Eo%2Bk%2A180%5Eo with k=%220+%2C+or+1%22 , meaning theta=157.5%5Eo or theta=337.5%5Eo .
Approximate values for sine and cosine of those angles are
,
so we could write the answers as
root%284%2C2%29%28cos%28157.5%5Eo%29%2Bi%2Asin%28157.5%5Eo%29%29=approximately-1.0987%2Bi%2A0.4551 , and
root%284%2C2%29%28cos%28337.5%5Eo%29%2Bi%2Asin%28337.5%5Eo%29%29=approximately1.0987-i%2A0.4551 .

4) 9i=9%28cos%2890%5Eo%29%2Bsin%2890%5Eo%29%29 and 1%2Bi=sqrt%282%29%28cos%2845%5Eo%29%2Bi%2Asin%2845%5Eo%29%29 ,
so their ratio can be calculated as
%289i%29%2F%281%2Bi%29%22=%22%22=%22%289%2Fsqrt%282%29%29%28cos%2890%5Eo-45%5Eo%29%2Bi%2Asin%2890%5Eo-45%5Eo%29%29%22=%22%289sqrt%282%29%2F2%29%28cos%2845%5Eo%29%2Bi%2Asin%2845%5Eo%29%29
Another way to calculate that quotient is using the conjugate of the denominator:
%289i%29%2F%281%2Bi%29%22=%22%289i%2F%281%2Bi%29%29%28%281-i%29%2F%281-i%29%29%22=%229i%281-i%29%2F%28%281%2Bi%29%281-i%29%29%22=%229%28i-i%5E2%29%2F%281%5E2-i%5E2%29=9%28i%2B1%29%2F%281%2B1%29%22=%229%281%2Bi%29%2F2=9%2F2%2Bi%2A%289%2F2%29%22=%22%289sqrt%282%29%2F2%29%28cos%2845%5Eo%29%2Bi%2Asin%2845%5Eo%29%29 .
Then,
%28%289i%29%2F%281%2Bi%29%29%5E%221+%2F+6%22%22=%22%289sqrt%282%29%2F2%29%5E%221+%2F+6%22%28cos%28theta%29%2Bi%2Asin%28theta%29%29 ,
where 6theta=45%5Eo%2Bk%2A360%5Eo with k=%220+%2C+1+%2C+2+%2C+3%2C+4%2C+or+5%22 .
That means
theta%22=%22%2845%5Eo%2Bk%2A360%5Eo%29%2F6%22=%227.5%5Eo%2Bk%2A60%5Eo .
%289sqrt%282%29%2F2%29%5E%221+%2F+6%22=approximately1.3613 .
Approximate values for sine and cosine of those angles are

So, the six complex values of %28%289i%29%2F%281%2Bi%29%29%5E%221+%2F+6%22 are
%289sqrt%282%29%2F2%29%28cos%287.5%5Eo%29%2Bi%2Asin%287.5%5Eo%29%29=approximately1.3497%2Bi%2A0.1777 ,
%289sqrt%282%29%2F2%29%28cos%2867.5%5Eo%29%2Bi%2Asin%2867.5%5Eo%29%29=approximately0.5209%2Bi%2A1.2577 ,
%289sqrt%282%29%2F2%29%28cos%28127.5%5Eo%29%2Bi%2Asin%28127.5%5Eo%29%29=approximately-0.8287%2Bi%2A1.0800 ,
%289sqrt%282%29%2F2%29%28cos%28187.5%5Eo%29%2Bi%2Asin%28187.5%5Eo%29%29=approximately-1.3497-i%2A0.1777 ,
%289sqrt%282%29%2F2%29%28cos%28247.5%5Eo%29%2Bi%2Asin%28247.5%5Eo%29%29=approximately-0.5209-i%2A1.2577 , and
%289sqrt%282%29%2F2%29%28cos%28307.5%5Eo%29%2Bi%2Asin%28307.5%5Eo%29%29=approximately0.8287-i%2A1.0800 .