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3 i. Since the equation
The equation z 2 2z 4 0 says that z 1 2 3 which yields z 1
also implies that z 2 z 2 2z 4 0 which says that z 3 8 0, the solutions of the equation
must be cube roots of 8.
3. Given two complex numbers z and w with w
similar identity for absolute value. 0, prove that the complex conjugate of z/w is z/w. Establish a Hint: Once the corresponding assertion for multiplication has been obtained we can obtain the
assertion for division very quickly using the fact that
z
z w w.
4. Work out 1 i n for a few positive integers n. What is 1 i 4 ? 5. Prove that if n is an even natural number that is not a multiple of 4 then the real part of 1 i n is zero. Hint:
Use mathematical induction.
We want to prove that whenever n is a positive integer, the real part of the number 1 i 4n 2 is
zero. This fact is obvious if n 1. Now given any positive integer n for which the real part of the
number 1 i 4n 2 is zero we have
1 i 4 n 1 2 1 i 4 n 2 1 i 4 4 1 i 4 n 2
and therefore this number has a zero real part. It follows from the principle of mathematical
induction that 1 i 4n 2 has a zero real part for every positive integer n.
6. Prove that if n is a natural number then 1 i n is real and positive if and only if n is a multiple of 8.
If n is any positive integer then, since
1 i 4n 1 i 4n 2 1 i 2 2i 1 i 4n 2
and since the number 1 i 4n 2 is imaginary, we deduce that 1 i 4n is real. We deduce that if n is
any even positive integer then the number 1 i n must either be real or imaginary and it follows
that the number
1 i n 1 1 i n i 1 i n
cannot be real. We conclude that if n is an odd positive integer then 1 i n cannot be real.
Finally, if n is a positive integer then since
n
1 i 4n 1 i 4
4n
we see that the number 1 i 4n is positive if and only if n is even. Therefore the number 1 i
real and positive if and only if n is a multiple of 8. n is 7. Assuming the standard identities from trigonometry, prove that if and are any real numbers then we have
cos i sin cos i sin cos i sin .
Then prove by induction (or otherwise) that if n is any natural number then
cos i sin n cos n i sin n. 93 Does the same identity hold if n is a negative integer?
The first equation follows simply:
cos i sin cos i sin cos cos sin sin i sin cos sin cos cos i sin .
The equation
cos i sin n cos n i sin n
is obvious when n 1. Now given any positive integer n for which the equation
cos i sin n cos n i sin n
holds we have
cos i sin n1 cos i sin n cos i sin
cos n i sin n cos i sin
cos n i sin n
and so it follows from the principle of mathematical induction that the equation
cos i sin n cos n i sin n
holds for every positive integer n. Since
cos i sin
1
cos i sin 1
cos i sin
cos i sin...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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