# Thread: simplifying complex number to rcisΘ form?

1. ## simplifying complex number to rcisΘ form?

Hey guys i'm hoping someone can help me with with problem.
i found z=-2+2i in rcisΘ (r.Θ) being (2√2, 3π/4)

I need to find w= √2cis 45 find zw in rcisΘ form and the reciprocal of z in a+bi form.

so far i got (1+1i) for w but i don't know if that is right.

Any help is appreciated!

2. ## Re: simplifying complex number to rcisΘ form?

Do you know that if \displaystyle \displaystyle \begin{align*} z_1 = r_1\,\textrm{cis}\,{(\theta _1)} \end{align*} and \displaystyle \displaystyle \begin{align*} z_2 = r_2\,\textrm{cis}\,{(\theta _2)} \end{align*}, then \displaystyle \displaystyle \begin{align*} z_1z_2 = r_1r_2\,\textrm{cis}\,{ \left( \theta_1 + \theta_2 \right) } \end{align*}?

3. ## Re: simplifying complex number to rcisΘ form?

no i did not know that.

4. ## Re: simplifying complex number to rcisΘ form?

would that be like
√2 x 2√2 (cos (π/4 + 3π/4) + isin (π/4 + 3π/4)

5. ## Re: simplifying complex number to rcisΘ form?

Assuming you meant to close that last bracket, then yes.

6. ## Re: simplifying complex number to rcisΘ form?

Awesome. Thanks so much!