Asap need help with part ii (1 Viewer)

cossine · Dec 3, 2021

amdspotter said:
View attachment 34390

Can you give the solution for i)?

CM_Tutor · Dec 3, 2021

$\begin{align*} wz &= \cfrac{3+4i}{5} \times \cfrac{5 + 12i}{13} \\ &= \left(\cfrac{3}{5} \times \cfrac{5}{13} + \cfrac{4i}{5} \times \cfrac{12i}{13}\right) + \left(\cfrac{3}{5} \times \cfrac{12i}{13} + \cfrac{4i}{5} \times \cfrac{5}{13}\right) \\ &= \cfrac{15 + 48i^2}{65} + \cfrac{(36 +20)i}{65} \\ &= -\cfrac{33}{65} + \cfrac{56i}{65} \end{align*}$

$\begin{align*} \text{Now, since $|z| = 1$ and $|w| = 1$, we can conclude that} \qquad |zw| &= |z||w| = 1^2 = 1 \\ \text{And thus,} \qquad \left|-\cfrac{33}{65} + \cfrac{56i}{65}\right| &= 1 \\ \sqrt{\left(\cfrac{33}{65}\right)^2 + \left(\cfrac{56}{65}\right)^2} &= 1 \\ \cfrac{33^2}{65^2} + \cfrac{56^2}{65^2} &= 1 \\ 33^2 + 56^2 &= 65^2 \end{align*}$

So, $a = 33$ and $b = 56$ (or vice versa) is one solution.

Note that we can check this solution by noting that

$\begin{align*} 65^2 - 56^2 &= (65 - 56)(65 + 56) \\ &= 9 \times 121 \\ &= (3 \times 11)^2 \\ &= 33^2 \\ \text{Thus confirming that} \qquad 65^2 &= 33^2 + 56^2 \end{align*}$

Paradoxica · Dec 3, 2021

$\begin{align*} w\bar{z} &= \cfrac{3+4i}{5} \times \cfrac{5 - 12i}{13} \\ &= \left(\cfrac{3}{5} \times \cfrac{5}{13} - \cfrac{4i}{5} \times \cfrac{12i}{13}\right) + \left(\cfrac{4i}{5} \times \cfrac{5}{13} - \cfrac{3}{5} \times \cfrac{12i}{13}\right) \\ &= \cfrac{15 - 48i^2}{65} + \cfrac{(20-36)i}{65} \\ &= \cfrac{63}{65} - \cfrac{16i}{65} \end{align*}$

So, $a = 63$ and $b = 16$ (or vice versa) is the other solution.

amdspotter · Dec 3, 2021

Paradoxica said:
$\begin{align*} w\bar{z} &= \cfrac{3+4i}{5} \times \cfrac{5 - 12i}{13} \\ &= \left(\cfrac{3}{5} \times \cfrac{5}{13} - \cfrac{4i}{5} \times \cfrac{12i}{13}\right) + \left(\cfrac{4i}{5} \times \cfrac{5}{13} - \cfrac{3}{5} \times \cfrac{12i}{13}\right) \\ &= \cfrac{15 - 48i^2}{65} + \cfrac{(20-36)i}{65} \\ &= \cfrac{63}{65} - \cfrac{16i}{65} \end{align*}$

So, $a = 63$ and $b = 16$ (or vice versa) is the other solution.

ah ok this actually makes sense thanks

amdspotter · Dec 3, 2021

CM_Tutor said:
$\begin{align*} wz &= \cfrac{3+4i}{5} \times \cfrac{5 + 12i}{13} \\ &= \left(\cfrac{3}{5} \times \cfrac{5}{13} + \cfrac{4i}{5} \times \cfrac{12i}{13}\right) + \left(\cfrac{3}{5} \times \cfrac{12i}{13} + \cfrac{4i}{5} \times \cfrac{5}{13}\right) \\ &= \cfrac{15 + 48i^2}{65} + \cfrac{(36 +20)i}{65} \\ &= -\cfrac{33}{65} + \cfrac{56i}{65} \end{align*}$

$\begin{align*} \text{Now, since $|z| = 1$ and $|w| = 1$, we can conclude that} \qquad |zw| &= |z||w| = 1^2 = 1 \\ \text{And thus,} \qquad \left|-\cfrac{33}{65} + \cfrac{56i}{65}\right| &= 1 \\ \sqrt{\left(\cfrac{33}{65}\right)^2 + \left(\cfrac{56}{65}\right)^2} &= 1 \\ \cfrac{33^2}{65^2} + \cfrac{56^2}{65^2} &= 1 \\ 33^2 + 56^2 &= 65^2 \end{align*}$

So, $a = 33$ and $b = 56$ (or vice versa) is one solution.

Note that we can check this solution by noting that

$\begin{align*} 65^2 - 56^2 &= (65 - 56)(65 + 56) \\ &= 9 \times 121 \\ &= (3 \times 11)^2 \\ &= 33^2 \\ \text{Thus confirming that} \qquad 65^2 &= 33^2 + 56^2 \end{align*}$

thanks

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Asap need help with part ii (1 Viewer)

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cossine

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Paradoxica

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