If z is unimodular complex number then z- 1-ia-1-ia4 has

The correct option is C 4 real roots #10; #10; #10; #9; #10; #9; #10; #9; #10; #9; #10; #10; #10;We have #10; (1+ia1 ia)^4 = z ...

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Standard XII

Mathematics

Purely Imaginary

If z is unimo...

Question

If z is unimodular complex number then $z = (\frac{1 + i a}{1 - i a})^{4}$ has

2 real 2 imaginary roots

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4 real roots

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4 imaginary roots

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3 real and imaginary roots

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Solution

The correct option is C 4 real roots

We have $(\frac{1 + i a}{1 - i a})^{4} = z$ is a fixed complex number
$\Rightarrow (\frac{1 + i a}{1 - i a})^{4} = c o s θ + i s i n θ$
$\frac{1 + i a}{1 - i a} = c o s (\frac{2 n π + θ}{4})$ $+ i s i n \frac{2 n π + θ}{4}$
$\Rightarrow \frac{1 + i a}{1 - i a} = c o s θ + i s i n θ$
Where, $ϕ = \frac{2 n ϕ + ϕ}{4}$
applying componendo dividend,
$\frac{1 + i a + 1 - i a}{1 + i a = 1 + i a} = \frac{c o s θ + i s i n ϕ + 1}{c o s θ + i s i n ϕ - 1}$
$\Rightarrow \frac{2}{2 i a} = \frac{c o s θ + i s i n ϕ + 1}{c o s θ + i s i n ϕ - 1}$
$\Rightarrow \frac{1}{i a} = \frac{2 ω s^{2} \frac{ϕ}{2} + 2 i s i n \frac{ϕ}{2} cos \frac{ϕ}{2}}{- 2 s i n^{2} \frac{ϕ}{2} + 2 i s i n \frac{ϕ}{2} cos \frac{ϕ}{2}}$
$= c o + \frac{θ}{2} \frac{(ω s \frac{θ}{2} + i s i n \frac{θ}{2})}{(- s i n \frac{θ}{2} + i s i n \frac{θ}{2})}$
$\Rightarrow a = - t a n \frac{θ}{2}$
$= - t a n \frac{2 n π + θ}{8}$
$n = 0, 1, 2, 3$
$i . e a = - t a n \frac{θ}{8}, - t a n (\frac{π}{4} + \frac{θ}{8}),$ $- t a n (\frac{π}{2} + \frac{θ}{8}),$
$4 r e a l r o o t s, - t a n (\frac{3 π}{2} + \frac{θ}{8}) \frac{1}{i a}$
$= i c o t \frac{θ}{2}$

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If z is unimodular complex number then z=(1+ia1−ia)4 has

If z is unimodular complex number then $z = (\frac{1 + i a}{1 - i a})^{4}$ has