Subtracting complex in polar

Given w =√2(cosine (p/4) + i sine (pi/4) ) and
z = 2 (cosine (pi/2) + i sine (pi/2) ),

what is w - z expressed in polar form?

Correct answer:

m =  1.4142
A =  -0.7854 rad

Step-by-step explanation:

$$\begin{gathered} w_x=\sqrt{2}\cdot \cos (\pi \mathrm{/}4)=\sqrt{2}\cdot \cos (3.1416\mathrm{/}4)=1 \\ {w}_y=\sqrt{2}\cdot \sin (\pi \mathrm{/}4)=\sqrt{2}\cdot \sin (3.1416\mathrm{/}4)=1 \\ {z}_x=2\cdot \cos (\pi \mathrm{/}2)=2\cdot \cos (3.1416\mathrm{/}2)\doteq 1.2246\cdot 10^{-16} \\ {z}_y=2\cdot \sin (\pi \mathrm{/}2)=2\cdot \sin (3.1416\mathrm{/}2)=2 \\ x=w_x-z_x=1-1.2246\cdot 10^{-16}=1 \\ y=w_y-z_y=1-2=-1 \\ m=\sqrt{x^2+y^2}=\sqrt{1^2+(-1{)}^2}=\sqrt{2}=1.4142 \end{gathered}$$

$$\begin{gathered} A=\arctan (y\mathrm{/}x)=\arctan ((-1)\mathrm{/}1)\doteq -0.7854\text{ rad } \\ \alpha =A\to \text{ °}=A\cdot {\displaystyle \frac{180}{\pi }}\text{ °}=(-0.7854)\cdot {\displaystyle \frac{180}{\pi }}\text{ °}=-45\text{ °} \end{gathered}$$

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Showing 1 comment:

Dr Math

please checkout our complex number / imaginary numbers calculator:

https://www.hackmath.net/en/calculator/complex-number?input=sqrt%282%29%2A%20%28cos%20%28pi%2F4%29%20%2B%20i%2A%20sin%20%28pi%2F4%29%20%29%20-%202%2A%28cos%20%28pi%2F2%29%20%2B%20i%2A%20sin%20%28pi%2F2%29%29

3 months ago  Like

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