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2i^4−5i^3+3i^2+√−81 rewritten in a+bi form?

User Syscall
by
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1 Answer

0 votes

Answer:

-1 + 14i.

Explanation:

i^2 = -1.

i^4 = i^2 * i^2

= -1 * -1 = 1.

i^3 = i^2 * i = -i.

√−81 = i√81 = 9i.

So

2i^4−5i^3+3i^2+√−81

= 2 - (5 * -i) - 3 + 9i

= -1 + 5i + 9i

= -1 + 14i.

User Rudik
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