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```        z^10-z^5-992=0 find number of roots of equation where real parts are negative
```
8 years ago

mycroft holmes
266 Points
```										z10-z5-992 =(z5-32) (z5+31) = 0
The roots of z5-32 = 0 are

. You can see that here two roots will have negative real parts.

This also tells us that z5+32 = 0 will have three such roots.

So, we have a grand total of 5.

```
8 years ago
Pratham Ashish
17 Points
```										Dear Piyush,
first assume z^5 = t ,
then eqn becomes t2-t -992 = 0
roots are t=32,-31
1st case t = 32 ;
which means z^5 = 32 , here u can see that |z| = 2 and since this eqn is of 5th power then it will hav 5 roots with |z|= 2 and five different argument angles with 72 as a gap b/w them .i.e, 360/5=72
now draw these roots on the complex plane starting from arg=0 , then 72, 144,216,288and 360.u will get 4 imaginary roots in which 2 will lie on the left side of the plane (i.e. with negative real part , they're of arg = 144 and 216) . mod of all the 5 roots will be 2.
now in 2nd case
similar to the above one , we will hav 5 roots |z| = 1.999 , but in this case if you draw 5 roots on the imaginary plane then u will find almost similar diagram with 3 on the left side this time cauz here eqn is z^5 = -31.
thus the total no. of roots with negative real parts is 5.
```
8 years ago
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