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Harmonic Progression

If z=2-3i, ...

Question

If $z = 2 - 3 i$ , show that $z^{2} - 4 z + 13 = 0$ and hence find the value of $4 z^{3} - 3 z^{2} + 169$ .

A

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B

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C

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Solution

The correct option is A 0
$z = 2 - 3 i$
$(z - 2) = - 3 i$
$(z - 2)^{2} = (- 3 i)^{2}$
$z^{2} - 4 z + 4 = 9 i^{2}$
$z^{2} - 4 z + 13 = 0$
Therefore,
$4 z^{3} - 3 z^{2} + 169 = 4 z (z^{2} - 4 z + 13) + 13 (z^{2} - 4 z + 13) = 4 z (0) + 13 (0) = 0$

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Similar questions

z = 2 - 3 i

, then show that,

z^{2} - 4 z + 13 = 0

z = 2 - 3 i

z^{2} - 4 z + 13 =

| z_{1} | = 2, | z_{2} | = 3, | z_{3} | = 4

| {2 z}_{1} + 3 z_{2} + {4 z}_{3} | = 4

then the absolute value of

{8 z}_{3} z_{2} + {27 z}_{3} z_{1} + {64 z}_{1} z_{2}

| z_{1} | = 2, | z_{2} | = 3, | z_{3} | = 4

| 2 z_{1} + 3 z_{2} + 4 z_{3} | = 0

, then absolute value of

8 z_{2} z_{2} + 27 z_{3} z_{1} + 64 z_{1} z_{2}

must be equal to

z_{1} \neq 0

z_{2}

be two complex numbers such that

\frac{z_{2}}{z_{1}}

is a purely imaginary number, then the value of

∣ ∣ ∣ \frac{2 z_{1} + 3 z_{2}}{2 z_{1} - 3 z_{2}} ∣ ∣ ∣

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