(a) Solve for Z in the equation $$ \mathrm{Z}^{4}=7 \mathrm{i} $$ expressing your answer in the cartesian form. (b) Given that $$ \mathrm{Z}_{1}=3+i, \mathrm{Z}_{2}=2+5 i $$ and $$ \frac{1}{\mathrm{Z}_{3}}=\frac{1}{\mathrm{Z}_{1}}+\frac{1}{\mathrm{Z}_{2}} $$ (i) Determine $$ \mathrm{Z}_{3} $$ in the form a +bi ; (ii) Represent $$ Z_{1}, Z_{2} $$ an $$ Z_{3} $$ on an Argand diagram.

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(a) To solve for $$ \mathrm{Z} $$ in the equation $$ \mathrm{Z}^{4}=7 \mathrm{i} $$, we first express $$ 7 \mathrm{i} $$ in polar form. Since $$ 7 \mathrm{i} $$ is a purely imaginary number, its magnitude is 7 and its angle is $$ \frac{\pi}{2} $$ radians.

So, $$ 7 \mathrm{i} = 7 \mathrm{e}^{\frac{\pi}{2} \mathrm{i}} $$.

Now, we take the fourth root of both sides:

$$ \mathrm{Z} = \left(7 \mathrm{e}^{\frac{\pi}{2} \mathrm{i}}ight)^{\frac{1}{4}} $$.

The fourth root of a complex number has four distinct values, each differing by a multiple of $$ \frac{\pi}{2} $$ radians in the argument. Therefore, we have:

$$ \mathrm{Z} = \sqrt[4]{7} \mathrm{e}^{\frac{\pi}{8} \mathrm{i}} $$, $$ \mathrm{Z} = \sqrt[4]{7} \mathrm{e}^{\frac{3\pi}{8} \mathrm{i}} $$, $$ \mathrm{Z} = \sqrt[4]{7} \mathrm{e}^{\frac{5\pi}{8} \mathrm{i}} $$, $$ \mathrm{Z} = \sqrt[4]{7} \mathrm{e}^{\frac{7\pi}{8} \mathrm{i}} $$.

Converting these to Cartesian form, we get:

$$ \mathrm{Z} = \sqrt[4]{7} \left(\cos\left(\frac{\pi}{8}ight) + \mathrm{i}\sin\left(\frac{\pi}{8}ight)ight) $$,
$$ \mathrm{Z} = \sqrt[4]{7} \left(\cos\left(\frac{3\pi}{8}ight) + \mathrm{i}\sin\left(\frac{3\pi}{8}ight)ight) $$,
$$ \mathrm{Z} = \sqrt[4]{7} \left(\cos\left(\frac{5\pi}{8}ight) + \mathrm{i}\sin\left(\frac{5\pi}{8}ight)ight) $$,
$$ \mathrm{Z} = \sqrt[4]{7} \left(\cos\left(\frac{7\pi}{8}ight) + \mathrm{i}\sin\left(\frac{7\pi}{8}ight)ight) $$.

(b) (i) To find $$ \mathrm{Z}_{3} $$, we first find the reciprocals of $$ \mathrm{Z}_{1} $$ and $$ \mathrm{Z}_{2} $$:

$$ \frac{1}{\mathrm{Z}_{1}} = \frac{1}{3+i} \cdot \frac{3-i}{3-i} = \frac{3-i}{10} $$,
$$ \frac{1}{\mathrm{Z}_{2}} = \frac{1}{2+5i} \cdot \frac{2-5i}{2-5i} = \frac{2-5i}{29} $$.

Now, we add these reciprocals:

$$ \frac{1}{\mathrm{Z}_{3}} = \frac{3-i}{10} + \frac{2-5i}{29} $$.

To add these fractions, we need a common denominator, which is $$ 10 	imes 29 = 290 $$:

$$ \frac{1}{\mathrm{Z}_{3}} = \frac{29(3-i) + 10(2-5i)}{290} $$,
$$ \frac{1}{\mathrm{Z}_{3}} = \frac{87 - 29i + 20 - 50i}{290} $$,
$$ \frac{1}{\mathrm{Z}_{3}} = \frac{107 - 79i}{290} $$.

Now, we take the reciprocal of both sides to find $$ \mathrm{Z}_{3} $$:

$$ \mathrm{Z}_{3} = \frac{290}{107 - 79i} $$.

To convert this to Cartesian form, we multiply the numerator and denominator by the conjugate of the denominator:

$$ \mathrm{Z}_{3} = \frac{290(107 + 79i)}{(107 - 79i)(107 + 79i)} $$,
$$ \mathrm{Z}_{3} = \frac{290(107 + 79i)}{107^2 + 79^2} $$,
$$ \mathrm{Z}_{3} = \frac{290(107 + 79i)}
(a) The solutions for $$ \mathrm{Z} $$ are $$ \sqrt[4]{7} \left(\cos\left(\frac{\pi}{8}ight) + \mathrm{i}\sin\left(\frac{\pi}{8}ight)ight) $$, $$ \sqrt[4]{7} \left(\cos\left(\frac{3\pi}{8}ight) + \mathrm{i}\sin\left(\frac{3\pi}{8}ight)ight) $$, $$ \sqrt[4]{7} \left(\cos\left(\frac{5\pi}{8}ight) + \mathrm{i}\sin\left(\frac{5\pi}{8}ight)ight) $$, $$ \sqrt[4]{7} \left(\cos\left(\frac{7\pi}{8}ight) + \mathrm{i}\sin\left(\frac{7\pi}{8}ight)ight) $$. (b) (i) The value of $$ \mathrm{Z}_{3} $$ in Cartesian form is $$ \frac{290(107 + 79i)}{107^2 + 79^2} $$.

(a) Solve for Z in the equation expressing your answer in the cartesian form.
(b) Given that
and
(i) Determine in the form a +bi ;
(ii) Represent an on an Argand diagram.

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