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I'd really appreciate if someone can explain to me how do you convert iz to to its' polar form? The answer is rcis(90/theta) I don't know how they got there.

Same question with -z = rcis(180+theta)

Thank you in advance!

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    $\begingroup$ $i = \cos 90^\circ + i\sin 90^\circ$ and $z = r(\cos\theta + i\sin \theta)$. Generally, if $z_1 = r_1(\cos\theta_1 + i\sin \theta_2), z_2 = (\cos\theta_2 + i\sin \theta_2)$, then $z_1z_2 = r_1r_2(\cos(\theta_1+\theta_2) + i \sin(\theta_1+\theta_2))$. Hence $iz = r(\cos(90^\circ+\theta)+i\sin(90^\circ+\theta))$ and since $-1 = \cos 180^\circ + i\sin 180^\circ$, $-z = r(\cos(180^\circ+\theta)+i\sin(180^\circ+\theta))$ $\endgroup$ – user348749 Sep 2 '16 at 8:52
  • $\begingroup$ try to look at mathsisfun.com/polar-cartesian-coordinates.html $cis\alpha=\cos\alpha+i\sin\alpha$ $\endgroup$ – gbox Sep 2 '16 at 8:55
  • $\begingroup$ Muralidharan, thank you so much! How do I upvote your answer and close the post? $\endgroup$ – Royi Sep 2 '16 at 9:07
  • $\begingroup$ Comments can not be up-voted. $\endgroup$ – user348749 Sep 2 '16 at 10:12
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iz = i x z = (i x |z|e^iθ) = [e^(iπ/2)] [|z|e^(iθ)] =|z| [e^(iπ/2+iθ)] =|z| [cos(π/2+θ)+isin(π/2+θ)] = rcis(π/2+θ) [|z|=r] [cos(a)+isin(a)=cis(a)]

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