Question

i(4+3i)/(1-i)find a and b​

Answer 1

Answer:

$a=\frac{1}{2}\:and\:b=\frac{7}{2}$

Step-by-step explanation:

The given complex number is converted to standard form by multiplying both numerator and denominator by conjugate of the denominator

Given:

$\frac{4+3i}{1-i}=a+i\:b$​

$\implies\:\frac{4+3i}{1-i}*\frac{1+i}{1+i}=a+i\:b$​

$\implies\:\frac{(4+3i)(1+i)}{1^2-i^2}=a+i\:b$​

$\implies\:\frac{4+4i+3i+3i^2}{1+1}=a+i\:b$​

$\implies\:\frac{4+7i+3(-1)}{2}=a+i\:b$​

$\implies\:\frac{4+7i-3}{2}=a+i\:b$​

$\implies\:\frac{1+7i}{2}=a+i\:b$​

$\implies\:\frac{1}{2}+\frac{i\:7}{2}=a+i\:b$​

$\text{separating real and imaginary parts, we get}$

$a=\frac{1}{2}\:and\:b=\frac{7}{2}$

Answer 2

Answer:

a=-7/2 and b=1/2

Step-by-step explanation:

=(4i+3i^2)/(1-i)×(1+i)/(1+i)

=(4i+3i^2)×(1+i)/(1)^2-(i)^2

=4i+3i^2+4i^2+3i^3/1-i^2

=4i+7i^2+3i^3/1-i^2

=4i-7-3i/2

=i-7/2

=-7+i/2

=-7/2+i/2

=by equality of complex number

a=-7/2 b=1/2