JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

Dividing

f(z)

z-i

, we obtain the remainder

i

and dividing it by

z+i

, we get the remainder

1+i

. The remainder upon the division of

f(z)

z^2+1

, is

View answer and explanation

Correct option: B

\frac{1}{2}(i z+1)+i

Option B (

\frac{{1}}{{2}}(i z+1)+i

) is correct because it is the remainder obtained after dividing f(z) by

z^2 + 1

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z \neq \pm 1

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\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}