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prohojiy [21]
3 years ago
14

Express the quotient of z1 and z2 in standard form given that

frac%7B-%5Cpi%20%7D%7B4%7D%20%29%2Bisin%28%5Cfrac%7B-%5Cpi%20%7D%7B4%7D%20%29%5D" id="TexFormula1" title="z_{1} = -3[cos(\frac{-\pi }{4} )+isin(\frac{-\pi }{4} )]" alt="z_{1} = -3[cos(\frac{-\pi }{4} )+isin(\frac{-\pi }{4} )]" align="absmiddle" class="latex-formula"> and z_{2} = 2\sqrt{2} [cos(\frac{-\pi }{2} )+isin(\frac{-\pi }{2} )]

Mathematics
1 answer:
Lesechka [4]3 years ago
3 0

Answer:

Solution : -\frac{3}{4}-\frac{3}{4}i

Step-by-step explanation:

-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}

=-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right) ÷ 2\sqrt{2}\left(0-1\right)i

= 3\left(-\frac{\sqrt{2}i}{2}+\frac{\sqrt{2}}{2}\right) ÷ -2\sqrt{2}i

= \frac{3\left(1-i\right)}{\sqrt{2}}÷ 2\sqrt{2}i = -3-3i ÷ 4 = -\frac{3}{4}-\frac{3}{4}i

As you can see your solution is the last option.

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<u>
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