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docker41 [41]
3 years ago
10

Which is equivalent to the complex number i^32?

Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0

Answer:

where i^2 = -1 (minus one)

So, i^2 = -1

i^4 = (i^2)^2 = (-1)^2 = -1 X-1 = 1

like ;

i^32 = (i^2)^16 = (-1)^16 =  -1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X

     = 1

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Now, solving for the value of p.

If equals are added to both sides of inequality, the inequality remains unchanged.

Now, -9 + p < 15

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