Math, asked by zijanupija, 1 year ago

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)

Answers

Answered by aagnavaagna
3

ANSWER EXPLANATION: 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+16i3i+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.

Answered by vovesihato
0


(8−i3−2i)(3+2i3+2i)=24+16i−3+(−i)(2i)(32)+(2i)2

24+16i3i+29−(−4)=26+13i13

The final answer is B.


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