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Convert 3 – 3i to polar form.
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i dont know how to do this but look at the website it might help u;)
Let 3-3 i=r(cos theta+i sin theta) \[r ~\cos \theta=3\] \[r \sin \theta=-3\] square and add \[r^2\left( \cos ^2\theta+\iota \sin ^2\theta \right)=3^2+(-3)^2=9+9=18=9 \times 2,r=3\sqrt{2}\] cos is positive and sin is negative so \[\theta\] lies in fourth quadrant. divide\[\tan \theta=\frac{ r \sin \theta }{ r \cos \theta }=\frac{ -3 }{ 3 }=-1=- \tan 45=\tan \left( 360-45 \right) =\tan 315\] \[\theta =315\] now you can write.
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