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)
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srilukolluru:
sorry friend don't know
in a+ib a is real part b is imaginary thats y i mentioned
hey friend i am giving the answer
ok n tq
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.
(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.
tq
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