Mathematics: If [tex] rac{a + ib}{c + id} [/tex]is purely real complex number then prove that: ad=bc

If [tex] \frac{a + ib}{c + id} [/tex]is purely

kristen17diaz

14.03.2022

Rewrite the given number as

$\dfrac{a+ib}{c+id}=\dfrac{(a+ib)(c-id)}{(c+id)(c-id)}=\dfrac{ac+bd+i(bc-ad)}{c^2+d^2}$

If it's purely real, then the complex part should be 0, so that

$\dfrac{bc-ad}{c^2+d^2}=0\implies bc-ad=0\implies\boxed{ad-bc}$

as required.

0,0(0 оценок)

kmontanab00

16.02.2020

It would be: ⁹P₈ = 9! (9-8)! = 9!/1! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 = 3,62,880

In short, Your Answer would be 3,62,880

Hope this helps!

0,0(0 оценок)