HKDSE Mathematics Question Review (20240907)

Question:

Suggest 3 methods to show that when a+bi=0, where a and b are real numbers, then a=0=b.

Solution:

Method 1

When a+bc=0 where c is not a rational number, then a and bc can never offset each other, just as a sum of apple is not equal to a sum of orange due to incompatible number sets and fruit sets.

Since I cannot be expressed as a ratio of integers, it is not a rational number, so it is impossible that a = bi.

QED

Method 2

We use the method of contradiction that when a and b must be equal, a=/=bi：

Put a= i and b= -1,

then a+bi = i - i =0

But this contradicts the fact that a is real.

Put a= 1 and b= i ,

then a+bi = 1 - 1 =0

But this contradict the fact that b is real.

Put a= i^(3/2) and b= -i^(1/2),

then a+bi = i^(3/2) - i^(3/2) =0

But this contradict the fact that a and be are real.

QED

We use the method of contradiction that when a and b must be real , then a=/=bi

Method 3

By looking at Argand diagram:

Clearly a and bi are independent and of different dimensions, and the only solution set (a,b) is (0,0).

QED

Method 4 (For reference only)

By inspection, sin(x + π/4) = cos(π). This implies that sin(x) and cos(x) has phase difference π/4 in radian measures.

Then rewrite a+bi into polar form:

re^(ix) = rcos(x) + rsin(x) I

Then by dot product, for known phase angle:

(a) • (bi)=|a||bi|cos(π/4) = 0

Then the geometry of Argand diagram is established, so a=/=bi.

QED

re^(ix) = rcos(x) + rsin(x) i

Since i cannot be expressed as a ratio of integers,

By inspection, sin(x + π/4) = cos(x)