PLs Help Question ASAP (1 Viewer)

Modern4DaBois · Dec 5, 2021

Sup can anyone pls help with these two Qs?
Thanks!!!!

Screen Shot 2021-12-05 at 1.13.34 pm.png

jimmysmith560 · Dec 5, 2021

Would this help for Question 21?

$\text{Let }\angle OAC = \alpha\:\text{and}\:\angle OCB = \beta\:\text{then}$

$\dfrac{z}{|\overrightarrow{AO}|}e^{i\alpha}=\dfrac{1-z}{|\overrightarrow{AC}|}$

$e^{i\alpha} = \dfrac{|z|(1-z)}{z|1-z|}$

Similarly

$\dfrac{1}{|\overrightarrow{CO}|}e^{i\beta}=\dfrac{\dfrac{1}{z}-1}{|\overrightarrow{CB}|}$

$e^{i\beta} = \dfrac{|z|(1-z)}{z|1-z|}$

Since the angles lie between zero and 180 degrees then they must be equal.

ExtremelyBoredUser · Dec 5, 2021

Modern4DaBois said:
Sup can anyone pls help with these two Qs?
Thanks!!!!
View attachment 34435

Q.22

a) just draw a diagram and it should be clear. there's an algebraic proof for this as well.

b)

$|\frac{z^2-4}{z}| = 2$
$|z^2-4| = 2|z|$

$|z|^2 - 4 \leq 2|z|$
$|z|^2 - 2|z| - 4 \leq 0$

The maximum of value of |z| is when the quadratic is 0.

do quadratic formula and you should get

|z| = 1 +- sqrt(5)

hence max value must be 1 + sqrt(5).

Modern4DaBois · Dec 5, 2021

Thank you so much for your help guys!!!

Drongoski · Dec 6, 2021

$\text {If you draw the diagram, you can show triangle AOC and triangle COB are similar. Why?}\\ \\ \frac {1}{z} = \frac {1\times \bar z}{z \times \bar z} = \frac {\bar z}{|z|^2} \\ \\ \text { and } \bar z \text{ is a reflection of z on the x-axis so that } \angle AOC = \angle COB \\ \\ \text { and since OC = 1, } \frac {OA}{OC} = \frac {OC}{OB} \because \frac {|z|}{1} = \frac {1}{OB} = \frac {|z|}{1} \text { since } OB = |\frac {1}{z} | = \frac {1}{|z|} \\ \\ \therefore \triangle AOC ||| \triangle COB \implies \text { corresponding angles equal} \\ \\ \therefore \angle OAC = \angle OCB$

QED

PLs Help Question ASAP (1 Viewer)

Modern4DaBois

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jimmysmith560

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ExtremelyBoredUser

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Modern4DaBois

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Drongoski

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