i give out 5 star super easy. 3. The complex numbers are numbers of the form a +
ID: 2986127 • Letter: I
Question
i give out 5 star super easy.
3. The complex numbers are numbers of the form a + bi (that is, a + b*i) where a, b are real numbers and the imaginary number i satisfies i^2 = -1. Addition is (a + bi) + (c + di) = (a+c) + (b + d)i. Multiplication is based on multiplication of coefficients, the property that i^2 = -1, and the distributive laws. Thus:
(a + bi)*(c + di) = a*(c + di) + bi*(c + di)
= a*c + a*di + bi*c + bi*di;
= a*c + a*d*i + b*c*i + b*d * i*i
= (a*c - b*d) + (a*d + b*c)*i.
3.a) Find the multiplicative inverse of (3/4) + (1/5)i.
3.b) Show that every non-zero complex number a + bi has a multiplicative inverse by observing that (a+bi)(a-bi) = a^2 + b^2, which is zero only when a and b are both = 0. Once you do this, then you know that the set of complex numbers is a field.
Explore: Let's try to define the complex numbers a + bi where the coefficients a and b come from Z/2Z. Call the set {a + bi : a, b in Z/2Z} by Z/2Z[i].
3.c) How many elements are in Z/2Z[i]? Write them down.
3.d) Write down the multiplication table for Z/2Z[i].
3.e) Is Z/2Z[i] a field? Explain.
Explanation / Answer
3.
a)
(400/241)[(3/4)-(1/5)i] (you get the idea from part b, since (a+ib)(a-ib) = a^2+b^2, we have (a+ib)z = 1, where
z = [1/(a^2+b^2)](a-ib).
b)
(a+ib)(a-ib) = a.a+a.(-ib)+(ib).a+(ib)(-ib) = a^2 -i.ab + i.ab +b^2 = a^2+b^2.
c)
4,
0+0.i = 0
0+1.i = i
1+0.i = 1
1+1.i = 1+i
(d)
0 i 1 1+i
0 0 0 0 0
i 0 1 i 1+i
1 0 i 1 1+i
1+i 0 1+i 1+i 0
(e)
No,
since (1+i)(1+i) = 0 in Z/2Z(i) and 1+i is nonzero.
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