If iz3 + z2-z+ i = 0, then show that z = 1. |z| = %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please explain the "alternative solution" How did we write that equation (the one with reciprocals)?
Problem 3:
If iz3 +z2-z+ i = 0, then show that z =
|z| = 1.
L.H.S. = iz (z – i) - (z – i) = (z – i) (iz² – 1) = 0
Either z - i = 0 =z =i= z = 1
or iz?- 1= 0= iz? = 1= iz?( – iz?) =1 = (zz)? = 1= z|= 1.
Solution:
Alternative Solution:
Take complex conjugate of the given equation. We get
i
-i z +z? -z-i=0=>
1
+i = 0.
z3 z? z
1
Comparing it with the given equation, we find the z =
= |z| = 1.
- | IN
Transcribed Image Text:Problem 3: If iz3 +z2-z+ i = 0, then show that z = |z| = 1. L.H.S. = iz (z – i) - (z – i) = (z – i) (iz² – 1) = 0 Either z - i = 0 =z =i= z = 1 or iz?- 1= 0= iz? = 1= iz?( – iz?) =1 = (zz)? = 1= z|= 1. Solution: Alternative Solution: Take complex conjugate of the given equation. We get i -i z +z? -z-i=0=> 1 +i = 0. z3 z? z 1 Comparing it with the given equation, we find the z = = |z| = 1. - | IN
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