Question

Express the product of z1 and z2 in standard form given that rac%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 $z_{2} = 2\sqrt{2} [cos(\frac{-\pi }{2} )+isin(\frac{-\pi }{2} )]$

Answer 1

Answer:

Solution : 6 + 6i

Step-by-step explanation:

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

This is the expression we have to solve for. Now normally we could directly apply trivial identities and convert this into standard complex form, but as the expression is too large, it would be easier to convert into trigonometric form first ----- ( 1 )

( Multiply both expressions )

$-6\sqrt{2}\left[\cos \left(\frac{-\pi }{4}+\frac{-\pi \:\:\:}{2}\right)+i\sin \left(\frac{-\pi \:}{4}+\frac{-\pi \:\:}{2}\right)\right]$

( Simplify $\left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ for both $\cos \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ and $i\sin \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ )

$\left(\frac{-\pi }{4}+\frac{-\pi }{2}\right)$ = $\left(-\frac{3\pi }{4}\right)$

( Substitute )

$-6\sqrt{2}\left(\cos \left(-\frac{3\pi }{4}\right)+i\sin \left(-\frac{3\pi }{4}\right)\right)$

Now that we have this in trigonometric form, let's convert into standard form by applying the following identities ----- ( 2 )

sin(π / 4) = √2 / 2 = cos(π / 4)

( Substitute )

$-6\sqrt{2}\left(-\sqrt{2} / 2 -i\sqrt{2} / 2 )$

= $-6\sqrt{2}\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)$ = $-\frac{\left(-\sqrt{2}-\sqrt{2}i\right)\cdot \:6\sqrt{2}}{2}$

= $-3\sqrt{2}\left(-\sqrt{2}-\sqrt{2}i\right)$ = $-3\sqrt{2}\left(-\sqrt{2}\right)-\left(-3\sqrt{2}\right)\sqrt{2}i$

= $3\sqrt{2}\sqrt{2}+3\sqrt{2}\sqrt{2}i:\quad 6+6i$ - Therefore our solution is option a.