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4-1.Complex numbers
easy
જો $\frac{{5( - 8 + 6i)}}{{{{(1 + i)}^2}}} = a + ib$ તો$(a,\,b)$ = . . .
A
$(15, 20)$
B
$(20, 15)$
C
$( - 15, 20)$
D
એકપણ નહીં.
Solution
(a) $\frac{{5( – 8 + 6i)}}{{{{(1 + i)}^2}}} = a + ib$==> $\frac{{ – 40 + 30i}}{{2i}} = 15 + 20i = a + ib$
Equating real and imaginary parts, we get $a = 15$ and $b = 20$.
Standard 11
Mathematics
