Sina Sanaizadeh – aged 15, Hinde House Secondary School , Sheffield
Q1: In general, what would you need to add to a+ib to get a real number?
We have to use the inverse of 'ib' to eliminate it from the expression. Therefore, to get a real number, you would add c-ib to a + ib, so long as c ≠-a.
Q2: ....Or an imaginary number?
We have to use the inverse of 'a' to eliminate it from the expression. Therefore, to get an imaginary number, you would add -a+ic to a+ib, so long as c ≠-b.
Q3: In general, what would you need to multiply by a+ib to get a real number?
I used the 'difference of 2 squares' identity to help me approach this question (that is, (a+b)(a-b) = a2– b2), because the iab and –iab would cancel each other out and i2b2 would become a real number (that is, –b2).
A multiplication grid makes it easy to see how the ‘middle terms’ would cancel out.
x a ib
a
a2 iab
-ib -iab -i2b2 = -1 x b2 = -b2
The result is a2 – b2 which has no imaginary parts (i.e. it is a real number).
Q4: ....Or an imaginary number?
I had two methods to solve this problem.
Method 1: (a+ia)2
This works only if the coefficient of i is the same value as the real number.
I expanded it using the identity (a+b)2 = a2 + b2 + 2ab.
So, (a+ia)2 = a2 + i2a2 + i2a2
= a2 -a2 + i2a2
= i2a2 ïƒ an imaginary number
Method 2: (a+ib)(b+ia)
If I swap the place of a and b in the second complex number to get (a+ib)(b+ia) then the real numbers (ab and –ab) will cancel out, leaving only ia2 + ib2. This factorises to make i(a2 + b2), which is an imaginary number.
A multiplication grid shows how this happens:
x a ib
b
ab ib2
+ia ia2 i2ab = -1ab = -ab