Product of complex numbers in polar form

**jbwtucker** · July 18th 2011, 05:15 PM

Two complex numbers, $z_{1}$ and $z_{2}$ , are expressed in polar form.

$z_{1}=(\sqrt{2},10^{\circ})$
$z_{2} = (2\sqrt{2}, 15^{\circ})$

What is the product of $z_{1}$ and $z_{2}$ ?

A. $(3\sqrt{2},25^{\circ})$
B. $(4, 25^{\circ})$
C. $(3\sqrt{2},150^{\circ})$
D. $(4, 150^{\circ})$

**mr fantastic** · July 18th 2011, 07:48 PM

Originally Posted by jbwtucker

Two complex numbers, $z_{1}$ and $z_{2}$ , are expressed in polar form.

$z_{1}=(\sqrt{2},10^{\circ})$
$z_{2} = (2\sqrt{2}, 15^{\circ})$

What is the product of $z_{1}$ and $z_{2}$ ?

A. $(3\sqrt{2},25^{\circ})$
B. $(4, 25^{\circ})$
C. $(3\sqrt{2},150^{\circ})$
D. $(4, 150^{\circ})$

Have you reviewed the formula (undoubtedly in your class notes or textbook) that says how two multiply two complex numbers written in polar form? What does it say? Therefore ....?

**jbwtucker** · July 18th 2011, 08:13 PM

Not working from a textbook. Prepping for a state test that covers algebra, trig, pre-calc, calculus, statistics, and probability. The only materials provided by the state are a sample test and some study suggestions.

I was looking for how to multiply complex numbers, didn't add "in polar form" to my search. Googled that, as you suggested, and found that you multiply magnitudes and add angles. So...

$(2)(2\sqrt{2})=4$
$10^{\circ} + 15^{\circ}=25^{\circ}$

So the answer is $(4, 25^{\circ})$ , which is B.

Thanks for the nudge.

**Siron** · July 19th 2011, 12:21 AM

Correct!

**HallsofIvy** · July 19th 2011, 08:03 AM

In general, $$r_1e^{i\theta_1}$$ r_2e^{i\theta_2}$= (r_1r_2)e^{i(\theta_1+ \theta_2)}$

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