Be realistic

Let z=a+ib, a∈{1,2,3,4} and b∈{1,2,3,4,5,6}

f(z) = (a+ib)^2 - 2(a+ib) + 2 = a^2 - b^2 + 2aib - 2a - 2ib + 2 = (a^2 - b^2 - 2a - 2) + (2ab - 2b)i

If f(z) is real, (2ab - 2b) must be zero. Two cases possible: 1. b=0 and a is any number in the domain 2. a=1 and b is any number in the domain

1st case is not possible since 0 is not there in the domain of b. There are 6 possibilities for the first case ( a is fixed and b is any one of the 6 numbers in its domain) Total number of cases: n(a)*n(b) = 24

a/b = 6/24 = 1/4

a+b = 5

I don't know if this is the correct method to do it so if anyone else has a better method, please do let me know.

Okay, the set Re(z) has four elements and the set Img(z) has six. From these two sets, we create a complex number. So the number of possible complex numbers would be given by the following formula:

${4 \choose 1} \times {6 \choose 1} = 24$

Now, let us assume that:

$z = a + bi$

Then, the function $f(z) = z^2 - 2z + 2$ becomes:

$f(z) = (a+bi)^2 -2(a+bi) + 2$

$= a^2 + b^2i^2 + 2abi - 2a - 2bi + 2$

$= (a^2 - b^2 - 2a + 2) + 2b(a - 1)i$

Now, for $z$ to be completely real, $2b( a - 1)i = 0$

This gives us two solutions: $b = 0$ or $a = 1$

We don't have 0 in the set Img(z). Hence, we take $a= 1$

Once again, we make use of combinations i.e., ${6 \choose 1}$ to arrive at the conclusion that there are six complex numbers that have their real part as 1. Hence, the probability that the complex number picked has no imaginary part would be $\frac {6}{24} = \frac {1}{4}$

This gives us the fraction $\frac {a}{b}$ since 1 and 4 are both positive and coprime. Thus,

$a + b = 1 + 4 = 5$

f(z) can be reduced to the form f(z)= (z-1)^2 +1. for the square of a complex number to be real, it should contain only the imaginary part. hence, for z=a+bi to have only imaginary part, a should be equal to 1 to be eliminated by the -1 under the square. so , there is six probabilities for the function to be real. now taking all the choices(4x6) , the answer will be clear. 6/24=1/4

let z = x+iy, where x is the real part and y is the imaginary part of the complex number z. Then f(z) is real if imaginary part of f(z) is zero. This implies 2xy-2y=0 or 2xy = 2y, which implies x must be 1. Probability that x=1 is 1/4=a/b, a=1, b=4. Hence a+b = 5.

let z = x+iy solving the expression we get x^2 - y^2 + i2xy - 2x - i2y +2 it will be real if img part is 0 so 2xy - 2y = 0, so x = 1; total number of possible answers for expression are 6 4 = 24 number of real answers are 1 6 = 6; so a/b = 6/24 = 1/4; a+b = 5

1. I first observe the function $f(z) = z^{2} - 2z + 2$ . The only possible way to make $f(z)$ real is to make $z^{2} - 2z$ real too. In other words we have to eliminate all imaginary parts of $z^{2} -2z$ .
2. So now let $z = a + bi$ , where $a$ is $Re(z)$ , $b$ is $Img(z)$ and $i$ is imaginary unit, as defined before. Then substitute $z = a + bi$ to $z^{2} - 2z$ getting $(a + bi)^{2} - 2(a + bi) = a^{2} - b^{2} + 2abi - 2a - 2bi = a^{2} - b^{2} - 2a + 2abi - 2bi$ . Remember that we have to cancel out the imaginary part of $z^2 - 2z$ . It implies that $2abi - 2bi = 0$ . Simplifying this we get $a = 1$ .
3. Point 2 informs us that we can set $f(z)$ as real number if and only if $z$ 's real part equals $1$ . Since there is no any rule for $z$ 's imaginary part. Thus, we can compose 6 kinds of $z$ value and because of this we can also compose 6 kinds of $f(z)$ value.
4. Meanwhile, the total number of $f(z)$ value equals $4 \times 6 = 24$ . Then as the conclution, the probability that $f(z)$ is real $= \frac {6}{24} = \frac {1}{4}$ . And so the answer is $1+4=\boxed{5}$

Look the f(z) development: (a+bi)²-2(a+bi)+2 -> a²+2abi-b²-2a-2bi+2 If f(z) is real, then 2abi-2bi=0 -> a=1. When a=1, b can be {1,2,3,4,5,6}. Probability: 6/24 = 1/4. Then a+b=1+4=5.

Express $z$ as $a + bi$ , where $a$ and $b$ are the real and imaginary parts of $z$ , respectively. Now $f(z)$ becomes:

$(a + bi)^2 - 2(a + bi) + 2 = a^2 + 2abi - b^2 - 2a - 2bi + 2 = (a^2 - b^2 - 2a + 2) + (a - 1)2bi$

The only imaginary part of $f(z)$ is then $(a - 1)2bi$ .

For $f(z)$ to be real, $(a - 1)2bi = 0 \Rightarrow a = 1$ or $b = 0$ . However, $b$ , the imaginary part of $z$ , may not be 0, so we now know that $a = 1$ . There are 4 possible real parts for $z$ but only 1 works, so the probability is $\frac{1}{4}$ and the answer is $1 + 4 = \boxed{5}$ .