Evaluate the following complex numbers and leave your results in

Evaluate the following complex numbers and leave your results in polar form: (a) $$\displaystyle{5}\angle{30}^{{\circ}}{\left({6}-{j}{8}+{\frac{{{5}\angle{60}^{{\circ}}}}{{{2}+{j}}}}\right)}$$
(b) $$\displaystyle{\frac{{{\left({10}\angle{60}^{{\circ}}\right)}{\left({35}\angle-{50}^{{\circ}}\right)}}}{{{\left({2}+{j}{6}\right)}-{\left({5}+{j}\right)}}}}$$

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Stephanie Mann
Step 1
As a rule of thumb, if we add or subtract complex numbers, we will do that in rectangular form while multiplication, divison will be performed in polar form.
(a) $$\displaystyle{5}\angle{30}^{{\circ}}{\left({6}-{j}{8}+{\frac{{{3}\angle{60}^{{\circ}}}}{{{2}+{j}}}}\right)}={5}\angle{30}^{{\circ}}{\left({6}-{j}{8}+{\frac{{{3}\angle{60}^{{\circ}}}}{{{2.236}\angle{26.565}^{{\circ}}}}}\right)}$$
$$\displaystyle={5}\angle{30}^{{\circ}}{\left({6}-{j}{8}+{1.3416}\angle{33.43}^{{\circ}}\right)}$$
$$\displaystyle={5}\angle{30}^{{\circ}}{\left({6}-{j}{8}+{1.1196}+{j}{0.7392}\right)}$$
$$\displaystyle={5}\angle{30}^{{\circ}}{\left({7.13}-{j}{7.261}\right)}$$
$$\displaystyle={5}\angle{30}^{{\circ}}{\left({10.176}\angle-{45.52}^{{\circ}}\right)}$$
$$\displaystyle={50.88}\angle-{15.52}^{{\circ}}$$
Step 2
(b) $$\displaystyle{\frac{{{\left({10}\angle{60}^{{\circ}}\right)}{\left({35}\angle{60}^{{\circ}}\right)}}}{{{\left({2}+{j}{6}\right)}-{\left({5}+{j}\right)}}}}={\frac{{{\left({10}\angle{60}^{{\circ}}\right)}{\left({35}\angle-{50}^{{\circ}}\right)}}}{{-{3}+{j}{5}}}}$$
$$\displaystyle={\frac{{{\left({10}\angle{60}^{{\circ}}\right)}{\left({35}\angle-{50}^{{\circ}}\right)}}}{{{5.83}\angle{120.96}^{{\circ}}}}}$$
$$\displaystyle={60.02}\angle-{110.96}^{{\circ}}$$
Not exactly what youâ€™re looking for?
Lauren Fuller
a) $$\displaystyle{\left({5}\angle{30}^{{\circ}}\right)}{\left({6}-{j}{8}+{1.1197}+{j}{0.7392}\right)}={\left({5}\angle{30}^{{\circ}}\right)}{\left({7.13}-{j}{7.261}\right)}$$
$$\displaystyle={\left({5}\angle{30}^{{\circ}}\right)}{\left({10.176}\angle-{45.52}^{{\circ}}\right)}={50.88}\angle-{15.52}^{{\circ}}$$
b) $$\displaystyle{\frac{{{\left({10}\angle{60}^{{\circ}}\right)}{\left({35}\angle-{50}^{{\circ}}\right)}}}{{{\left(-{2}+{j}{5}\right)}={\left({5.83}\angle{120.96}^{{\circ}}\right)}}}}={60.02}\angle-{110.96}^{{\circ}}$$