Solve-in-C-the-equation-z-4-7-i-z-3-12-15i-z-2-4-4i-z-16-192i-0-Knowing-that-it-has-one-real-root-and-a-purely-imaginary-root-of-equal-magnitude-

Question Number 180274 by Ar Brandon last updated on 09/Nov/22

Solve in C the equation

z^{4} + (7 - i) z^{3} + (12 - 15 i) z^{2} + (4 + 4 i) z + 16 + 192 i = 0

Knowing that it has one real root and a purely imaginary root

of equal magnitude .

Answered by Frix last updated on 09/Nov/22

because at least one root is real we need

the imaginary part to equal 0

{\begin{cases} z^{4} + 7 z^{3} + 12 z^{2} + 4 z + 16 = 0 \\ i (z^{3} + 15 z^{2} - 4 z - 192) = 0 \Rightarrow z_{1} = - 4 (trying factors of 192) \end{cases}

\Rightarrow z_{2} = - 4 i \lor z_{2} = 4 i

trying in original equation we get

z_{2} = 4 i

\Rightarrow

z^{4} + (7 - i) z^{3} + (12 - 15 i) z^{2} + (4 + 4 i) z + 16 + 192 i =

= (z + 4) (z - 4 i) (z^{2} + (3 + 3 i) z - 12 + i)

\Rightarrow

z_{3} = - 5 - 2 i

z_{4} = 2 - i

Commented by Ar Brandon last updated on 10/Nov/22

Thanks. How did you get z3 and Z4 please?

Commented by Frix last updated on 10/Nov/22

you can use the usual formula

z^{2} + (3 + 3 i) z - 12 + i = 0

z = - \frac{3 + 3 i}{2} \pm \sqrt{\frac{{(3 + 3 i)}^{2}}{4} - (- 12 + i)} =

= - \frac{3 + 3 i}{2} \pm \sqrt{12 + \frac{7}{2} i}

solving {(a + b i)}^{2} = 12 + \frac{7}{2} i \Rightarrow \sqrt{12 + \frac{7}{2} i} = \frac{7 + i}{2}

z = - \frac{3 + 3 i}{2} \pm \frac{7 + i}{2} = {\begin{cases} - 5 - 2 i \\ 2 - i \end{cases}

Commented by Ar Brandon last updated on 10/Nov/22

Let me try

\Rightarrow z^{2} + (3 + 3 i) z - 12 + i = 0

\Rightarrow z = \frac{- 3 - 3 i \pm \sqrt{{(3 + 3 i)}^{2} + 4 (12 - i)}}{2}

= \frac{- 3 - 3 i \pm \sqrt{48 + 14 i}}{2} = \frac{- 3 - 3 i \pm (7 + i)}{2}

z_{3} = - 5 - 2 i, z_{4} = 2 - i

Commented by Ar Brandon last updated on 10/Nov/22

Got it. Thanks Sir !

Commented by Frix last updated on 10/Nov/22

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Commented by Ar Brandon last updated on 10/Nov/22

What about z^{2} + (3 + 3 i) z - 12 + i ?

Any quick methode to get it ? Appart from

{Euclid}^{'} s division .