Express the quotient of z1 and z2 in standard form given that [tex]z_{1} = 6[cos(\frac{2\pi }{5}) + isin(\frac{2\pi }{5})][/tex] and [tex]z_{2} = 2\sqrt{2} [cos(\frac{-\pi }{2}) + isin(\frac{-\pi }{2})][/tex]
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Answer:
Solution : - 2.017 + 0.656i
Step-by-step explanation:
The quotient of the two expressions would be the following,
[tex]6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right][/tex] ÷ [tex]2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right][/tex]
So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,
( 1 ) cos(x) = sin(π / 2 - x)
( 2 ) sin(x) = cos(π / 2 - x)
If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,
( 1 ) [tex]\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}[/tex]
( 2 ) [tex]\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}[/tex]
These two identities makes sin(π / 10) = [tex]\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}[/tex], and cos(π / 10) = [tex]\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}[/tex].
Therefore cos(2π / 5) = [tex]\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}[/tex], and sin(2π / 5) = [tex]\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}[/tex]. Substitute,
[tex]6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right][/tex] ÷ [tex]2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right][/tex]
Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,
[tex]6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right][/tex] ÷ [tex]2\sqrt{2}\left[0-i\right][/tex]
And now simplify this expression to receive our answer,
[tex]6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right][/tex] ÷ [tex]2\sqrt{2}\left[0-i\right][/tex] = [tex]-\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i[/tex],
[tex]-\frac{3\sqrt{5+\sqrt{5}}}{4}[/tex] = [tex]-2.01749\dots[/tex] and [tex]\:\frac{3\sqrt{3-\sqrt{5}}}{4}[/tex] = [tex]0.65552\dots[/tex]
= [tex]-2.01749+0.65552i[/tex]
As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.