# Complex numbers how to convert iz to its' polar form?

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!

• $i = \cos 90^\circ + i\sin 90^\circ$ 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 $-1 = \cos 180^\circ + i\sin 180^\circ$, $-z = r(\cos(180^\circ+\theta)+i\sin(180^\circ+\theta))$ – user348749 Sep 2 '16 at 8:52
• try to look at mathsisfun.com/polar-cartesian-coordinates.html $cis\alpha=\cos\alpha+i\sin\alpha$ – gbox Sep 2 '16 at 8:55
• Muralidharan, thank you so much! How do I upvote your answer and close the post? – Royi Sep 2 '16 at 9:07
• Comments can not be up-voted. – user348749 Sep 2 '16 at 10:12

## 1 Answer

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)]