- #1

- 565

- 2

The answer in the back of the book says its 2^16e(i8pie)

But z=|z|^(n)e(i(n)theta) Therefore the hypotenuse which is 2^(1/2) when multiplied by 23

should be 2^(23/2) not 2^(16)

- Thread starter Ry122
- Start date

- #1

- 565

- 2

The answer in the back of the book says its 2^16e(i8pie)

But z=|z|^(n)e(i(n)theta) Therefore the hypotenuse which is 2^(1/2) when multiplied by 23

should be 2^(23/2) not 2^(16)

- #2

- 737

- 0

You're right, of course. I don't know from where the book got 2^{16}.

- #3

tiny-tim

Science Advisor

Homework Helper

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Hey guys!Im trouble getting the correct answer for z^23 where z=1+1

The answer in the back of the book says its 2^16e(i*pie)

It's obviously z^32 …

(btw, if you type alt-p, it prints π)

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