8−i3−2i
If the expression above is rewritten in the form a+bi, where a and b are real numbers, what is the value of a? (Note: i=−1)
Anonymous:
Is your note correct?
why
i^2 = - 1. Isn't it?
a = 8
To rewrite 8−i3−2i in the standard form a+bi, you need to multiply the numerator and denominator of 8−i3−2i by the conjugate, 3+2i. This equals
(8−i3−2i)(3+2i3+2i)=24+16i−3+(−i)(2i)(32)+(2i)2
Since i2=−1, this last fraction can be reduced simplified to
24+16i−3i+29−(−4)=26+13i13
which simplifies further to 2+i. Therefore, when8−i3−2i is rewritten in the standard form a + bi, the value of a is 2.
The final answer is A.
(8−i3−2i)(3+2i3+2i)=24+16i−3+(−i)(2i)(32)+(2i)2
Since i2=−1, this last fraction can be reduced simplified to
24+16i−3i+29−(−4)=26+13i13
which simplifies further to 2+i. Therefore, when8−i3−2i is rewritten in the standard form a + bi, the value of a is 2.
The final answer is A.
Question should be like this 8-i / ( 3 - 2i )
Answers
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8-i3 (or) 3i-2i
8-5i=a+bi
by comparing both
a=-5i,b=8
given,i=-1
-5i=-5×-1
a=5
8-5i=a+bi
by comparing both
a=-5i,b=8
given,i=-1
-5i=-5×-1
a=5
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