Complex Mania-II
(
1
+
i
1
−
i
)
1
0
0
If the expression above equals
a
+
i
b
for real numbers
a
and
b
,then find the value of
a
+
b
.
The answer is 1.
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3 solutions
Moderator note:
That's right. Another way is to express 1 ± i = 2 ( 2 1 ± 2 i ) to polar coordinates then apply De Moivre's Theorem.
Bonus question : What would the answer be if we replace the number 100 by 99 instead?
1 + i 1 − i = ( 1 + i ) ( 1 − i ) ( 1 − i ) ( 1 − i ) = 1 − i 2 1 − 2 i + i 2 = 2 − 2 i = − i ⇒ ( − i ) 9 9 = ( − 1 ) i 3 = i ⇒ a + b = 1
In fact the possible values of a + b = { − 1 , 1 } . Because ( − i ) n = { − i , − 1 , i , 1 } congruence mod 4.
1 + i 1 − i = ( 1 + i ) ( 1 − i ) ( 1 − i ) ( 1 − i ) = 1 − i 2 1 + i 2 − 2 i
= 1 + 1 1 − 1 − 2 i = − i
( − i ) 1 0 0 = 1 (even power of any -ve number is +ve))
1 can be written as 1 + i 0 where a + b = 1 + 0 = 1
It'll be -2i instead of +2i. And hence the result will be -i. But yet it'll give the final result 1.
There's a mistake Siddharth Singh You wrote − i 1 0 0 = 1 .That is incorrect.Because: ( − a ) 2 n = a 2 n BUT − a 2 n = − 1 × a 2 n See what I'm talking about? You see, − i 1 0 0 = 1 because − i 1 0 0 = − 1 × i 1 0 0 = − 1 × 1 = − 1 . I 've corrected it.So, no need to worry.
First learn and then teach Siddharth Singh
