Porisma

Worked solution

Practise comparing components with real parameters

Find the real values of aa and bb given the following:

(a) (2+ai)(1+i)=5+ib(2+ai)(1+i)=5+ib.

(b) (a−3i)(1+ib)=−1−5i(a-3i)(1+ib)=-1-5i.

CJC · Self-Practice 2

(a) A linear pair

Answer

a=−3a=-3, b=−1b=-1.

Expand and collect terms:

(2+ai)(1+i)=2+2i+ai+ai2=(2−a)+(2+a)i.\begin{aligned}&(2+ai)(1+i)\\ & =2+2i+ai+ai^2\\ & =(2-a)+(2+a)i.\end{aligned}

Comparing with 5+ib5+ib gives 2−a=52-a=5 and 2+a=b2+a=b. Thus a=−3a=-3 and b=2−3=−1b=2-3=-1.

How did this part go?

(b) Two paired solutions

Answer

(a,b)=(−3,23)(a,b)=\left(-3,\frac23\right) or (2,−1)(2,-1).

Expand and use i2=−1i^2=-1:

(a−3i)(1+ib)=a+abi−3i−3bi2=(a+3b)+(ab−3)i.\begin{aligned}&(a-3i)(1+ib)\\ & =a+abi-3i-3bi^2\\ & =(a+3b)+(ab-3)i.\end{aligned}

Comparing components gives a+3b=−1a+3b=-1 and ab−3=−5ab-3=-5, so ab=−2ab=-2. Substitute a=−1−3ba=-1-3b:

(−1−3b)b=−2,3b2+b−2=0,(3b−2)(b+1)=0.\begin{aligned}(-1-3b)b&=-2,\\3b^2+b-2&=0,\\(3b-2)(b+1)&=0.\end{aligned}

For b=23b=\frac23, a=−3a=-3; for b=−1b=-1, a=2a=2. These are paired solutions, not four independent combinations.

How did this part go?

Solution notes

Both copies print the two questions and all answer pairs. The working here is authored from those printed equations; no source derivation is implied.