Question

find all the solutions of the equation z^5=(32/sqrt2)i-(32/sqrt2)

Answer 1

not sure if the question asking this solution, but the answer that i got his :
32^5[cos 3pi/4*(5) +i sin 3pi/4*(5)]

Answer 2

\[32(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)\]
where on the unit circle do we have (-1/sqrt(2),1/sqrt(2)) ?

Answer 3

for what angle?

Answer 4

-45 degrees (135degrees)

Answer 5

isn't it -,+ in the second quadrant

Answer 6

so it would be 135 degrees

Answer 7

yea exactly 135 would land on 3pi/4

Answer 8

which would be (-sqrt2/2,sqrt2/2)

Answer 9

\[z^5=32(\cos(135^o)+i \sin(135^o)) \\ \text{ we we can write } \\ z^5=32(\cos(135^o+360^o n)+i \sin(135^o+360^o n) )\]
So raise both sides to the 1/5 power
and since you want 5 roots
choose n=0 to get one 5th root
choose n=1 to get another 5th root
...
and
then finally
choose n=4 to get the last 5th root

Answer 10

soo that's what the questions means by finding all the solutions , i wouldn't just multiply 5 by3pi/4 for both cos and sin?

Answer 11

\[z=32^\frac{1}{5}(\cos(\frac{135^o+360^on}{5})+i \sin(\frac{135^o+360^on}{5}))\]

Answer 12

ok got it , you found the complex root (cos (theta/n)+(2kpi/n)......

Answer 13

thanks for your help