Question

Prove that equation 2z

5 + 8z – 1 = 0 has

one real root in | z | < 1 and four roots lie

between the circles | z | = 1, and | z | = 2.​

Answer 1

SOLUTION

TO PROVE

The equation $\sf{2 {z}^{5} + 8z - 1 = 0}$ has one real root in | z | < 1 and four roots lie between the circles | z | = 1, and | z | = 2.

EVALUATION

First suppose that

$\sf{f(z) = 2 {z}^{5} \: \: and \: \: \: g(z) = 8z - 1 }$

Then for | z | < 1

$\displaystyle \sf{ \frac{ |f(z)| }{ |g(z)| } }$

$\displaystyle \sf{ = \frac{ |2 {z}^{5} | }{ |8z - 1 | } }$

$\displaystyle \sf{ \leqslant \frac{ 2{| {z}^{} |}^{5} }{ |8z | - | 1 | } }$

$\displaystyle \sf{ \leqslant \frac{ 2 }{ 8 } }$

$\displaystyle \sf{ \leqslant 1 }$

So by Rouche's Theorem g(z) and f(z) + g(z) have same number of zeroes in | z | < 1

Since g(z) have one zero in | z | < 1

So f(z) + g(z) have one zero in | z | < 1

Next suppose that

$\sf{f(z) = 2 {z}^{5} \: \: and \: \: \: g(z) = 8z - 1 }$

Now

$\displaystyle \sf{ \frac{ |g(z)| }{ |f(z)| } }$

$\displaystyle \sf{ = \frac{ | 8z - 1 | }{|2 {z}^{5} |} }$

$\displaystyle \sf{ \leqslant \frac{ | 8z | + | 1 | }{2| {z}^{5} |} }$

$\displaystyle \sf{ \leqslant \frac{ 16 + 1 }{2 \times {2}^{5} } }$

$\displaystyle \sf{ \leqslant \frac{ 17 }{64} }$

$\displaystyle \sf{ \leqslant 1 }$

So by Rouche's Theorem f(z) and f(z) + g(z) have same number of zeroes in | z | < 2

Since f(z) have five zeroes in | z | < 2

So f(z) + g(z) have five zeroes in | z | < 2

But f(z) + g(z) have one zero in | z | < 1

So f(z) + g(z) have four zeroes in 1 < | z | < 2

Hence the equation $\sf{2 {z}^{5} + 8z - 1 = 0}$ has one real root in | z | < 1 and four roots lie between the circles | z | = 1, and | z | = 2.

Hence the proof follows

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