find the complex number which when multiplied by 5+3i shall give 3-4i

Question

plz tell me the way to solve such sums

Answer 1

Let the complex number be a+ib, then:

(a+ib)(5+3i)=3-4i.

1/(a+ib)=(a-ib)/(a²+b²), so 5+3i=(a-ib)(3-4i)/(a²+b²).

5+3i=(3a-i(4a+3b)-4b)/(a²+b²).

Equating real and imaginary parts:

(3a-4b)/(a²+b²)=5; (4a+3b)/(a²+b²)=-3.

3a-4b=5(a²+b²); 4a+3b=-3(a²+b²).

Therefore:

(3a-4b)/(4a+3b)=-5/3, 3(3a-4b)=-5(4a+3b),

9a-12b=-20a-15b, 3b=-29a, b=-29a/3, 

a²+b²=a²+841a²/9=850a²/9.

4a+3b=4a-29a=-25a=-3(850a²/9)=-850a²/3.

Divide through by -25a (since cannot be zero):

1=34a/3, a=3/34, b=-29a/3=-29/34.

So, the number is (3-29i)/34.

CHECK

(5+3i)(3-29i)/34=(15-136i+87)/34=(102-136i)/34=3-4i. OK