don't understand how phi(a+c)=a+c. Please explain. Thank you

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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I don't understand how phi(a+c)=a+c. Please explain. Thank you 

2:03
b)
The objective is to check whether the given
function : C→C is a ring homomorphism.
Step 5/20
Consider the given function : C→C which is
(a+bi) = a-bi
defined by
for all a, b in R
To check whether the given function is ring
homomorphism, it is enough to check the
following two ring homomorphism conditions for
all x, y in C
i) (x + y) = (x) + (y)
ii)(xy)=(x)o(y)
lll 22% €
Step 6/20
Check the first condition: $(x+y)= $(x)+ $(y)
for all x, y in C
Let x = a + bi and y=c+di, such that
(x) = a - bi and (y)=c-di
(x + y) = ((a+bi)+(c+di))
=p(a+c+(b+d)i)
= (a + c)-(b+d)i
Transcribed Image Text:2:03 b) The objective is to check whether the given function : C→C is a ring homomorphism. Step 5/20 Consider the given function : C→C which is (a+bi) = a-bi defined by for all a, b in R To check whether the given function is ring homomorphism, it is enough to check the following two ring homomorphism conditions for all x, y in C i) (x + y) = (x) + (y) ii)(xy)=(x)o(y) lll 22% € Step 6/20 Check the first condition: $(x+y)= $(x)+ $(y) for all x, y in C Let x = a + bi and y=c+di, such that (x) = a - bi and (y)=c-di (x + y) = ((a+bi)+(c+di)) =p(a+c+(b+d)i) = (a + c)-(b+d)i
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