Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
z^10-z^5-992=0 find number of roots of equation where real parts are negative
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.
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.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !