1. Find all complex roots of the equation z2+z-1-2iz-i=0.
z2+z-1-2iz-i = z2-2iz -1+z-i = (z-i)2+z-i = (z-i)(z-i+1). Hence z1 = i and z2 = -1+i.
2. Does the set {a+biC\{0} : a=0  b=0  a=b  a=-b} form a group under multiplication
of complex numbers?
The answer is YES. Let us denote the set by A. First we must show that A is closed under
multiplication. From the definition of the set A it follows that A = {r(cos+isin):=k/4,
k=0,1,…,7 and rR\{0}}. Given z1 and z2 from A we have z1= r1(cosk/4 +isink/4) and z2=
r2(cost/4+isint/4). Now z1z2 = r1r2(cos((k+t)mod8)/4 + sin((k+t)mod8)/4). Obviously
r1r20 and (k+t)mod8 {0,1, …, 7} so z1z2 belongs to A. Associativity is obvious, the identity
element for multiplication is 1+0i  A because its argument is 0. The inverse to r(cosk/4
+isink/4) is (1/r)(cos(8k)/4 +isin(8k)/4) that obviously belongs to A.
3. Group (G1,f1) is isomorphic to (G2,f2) and group (H1,s1) is isomorphic to (H2,s2). Does this
imply that G1H2 is isomorphic to H1G2 (with componentwise operations)?
YES. Suppose :G1G2 and :H2H1 be isomorphisms. We define : G1H2  H1G2 as
(a,b)=((b),(a)). Obviously  is a bijection. Let (x,y) and (p,q) be from G1H2. Denote the
operation in G1H2 by # and in H1G2 by *. Consider ((x,y)#(p,q)) = (xf1p,ys2q) =
((ys2q),(xf1p)) = ((y)s1(q),(x)f2(p))
because  and  are isomorphisms
= ((y),(x))*((q),(p))
by the definition of *
= (x,y)*(p,q)
by the definition of .
4. Define operations  and  on R+ in such a way that the function f(x)=ln x is an
isomorphism of the field (R+,,) with (R,+,) ( denotes regular multiplication).
First ln(a*b) = lna + lnb = lnab, hence a*b=ab. Then ln(ab) = lnalnb, hence ab=e(lnalnb)