Problems from Complex numbers in trigonometric form: product and quotient

Development:

${\displaystyle \frac{21\cdot [\cos ({225}^{\circ })+i\cdot \sin ({225}^{\circ })]}{9\cdot [\cos ({180}^{\circ })+i\cdot \sin ({180}^{\circ })]}}={\displaystyle \frac{27}{9}}\cdot [\cos ({225}^{\circ }-{180}^{\circ })+i\cdot \sin ({225}^{\circ }-{180}^{\circ })]$

$=3\cdot [\cos ({45}^{\circ })+i\cdot \sin ({45}^{\circ })]=3\cdot e^{i{45}^{\circ }}$

Solution:

$3\cdot e^{i{45}^{\circ }}$

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Development:

• First we change it to polar form $${\displaystyle z=3+3i\Rightarrow \left\{\begin{array}{l}|z|=\sqrt{3^2+3^2}=\sqrt{18}\\ \alpha =\arctan ({\displaystyle \frac{3}{3}})={45}^{\circ }\end{array}\right\}\Rightarrow z={\sqrt{18}}_{{45}^{\circ }}}$$

  And now we calculate the trigonometric form: $$z=\sqrt{18}\cdot [\cos ({45}^{\circ })+i\cdot \sin ({45}^{\circ })]=\sqrt{18}\cdot e^{i{45}^{\circ }}$$

• Since it is already written in polar form it is straight forward that: $$z=5\cdot [\cos ({180}^{\circ })+i\cdot \sin ({180}^{\circ })]=5\cdot e^{i{180}^{\circ }}$$

Solution:

• $z=\sqrt{18}\cdot e^{i{45}^{\circ }}$
• $z=5\cdot e^{i{180}^{\circ }}$

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