If - a - ib- - 1 then the simplified form of 1 - b - ai-1 - b - ai is

If $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ $A_{i}, B_{i}, C_{i}$ are respectively are cofator of $a_{i}, b_{i}, c_{i}$ -- for all i = 1, 2, 3, then

$∣ ∣ ∣ ∣ \begin{matrix} A_{1} & B_{1} & C_{1} A_{2} & B_{2} & C_{2} A_{3} & B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ ∣$

Q. If

A_{i} = [\begin{matrix} a^{i} & b^{i} b^{i} & a^{i} \end{matrix}]

and

| a | < 1, | b | < 1

then

\sum_{i = 1}^{\infty} | A_{i} |

is equal to

Q. The equation

z^{10} + (13 z - 1)^{10} = 0

has

5

pairs of complex roots

a_{1}, b_{1}, a_{2}, b_{2}, a_{3}, b_{3}, a_{4}, b_{4}, a_{5}, b_{5}

. If each pair

a_{i}, b_{i}

are complex conjugates, then

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If |a+ib|=1 then the simplified form of 1+b+ai1+b−ai is

The correct option is A b+ai Givem |a+ib|=1√a2+b2=1a2+b2=1b2=1−a2 Also, 1+b+ai1+b−ai=(1+b+ai)(1+b−ai)×(1+b+ai)(1+b+ai)=1+b2+a2(i)2+2b+2ia+2iab(1+b)2−(a)2(i)2=1+b2+a2+2b+2ia+2iab12+b2+2b+a2=2b2+2b+2ia(1+b)2+2b=2b(1+b)+2ia(1+b)2(1+b)=b+iaoption A is correct

If $| a + i b | = 1$ then the simplified form of $\frac{1 + b + a i}{1 + b - a i}$ is