Finding z in the form $a + bi$ when $z^2 = 1 + i + {58\over{9(3-7i)}}$

Question

This is a problem from Mathematics Analyses and Approaches HL (IB). Do note that this is not homework or any sort of submission, I'm doing it merely out of interest. This might be trivial for this website, but I've been stuck on it for about thirty minutes now.

The problem statement is as follows:

Given that: $$ z^2 = 1 + i + {58\over {9(3-7i)}} $$ find z in the form $ a + bi $ Here's what I've tried so far: $$ z^2 = {{9(3-7i)} \over {9(3-7i)}} + {{9i(3-7i)} \over {9(3-7i)}} + {58\over {9(3-7i)}} $$ $$ z^2 = { {9(3-7i) + 9i(3-7i) + 58}\over {9(3-7i)}} * {(3+7i)\over(3+7i)} $$ $$ z^2 = { {9(58) + 9i(58) + 58(3+7i)}\over {9(58)}} $$ $$ z^2 = { {9(58) + 9i(58) + 58(3+7i)}\over {9(58)}} $$ $$ z^2 = { {9 + 9i + 3+7i}\over 9} $$ $$ z^2 = {12 \over 9} + i{16 \over 9} $$ I tried taking z as $a + bi $, and then equating $a^2 - b^2$ to $12 \over 9$, and $2abi$ to $i{16 \over 9}$, but that doesn't seem to lead anywhere. Any help would be gladly appreciated.

Answer 1

Hint:

$$z^2 =\frac{12}{9}+\frac{16i}{9}=\frac{16}{9}+\frac{16\cdot i}{9}+\frac{4i^2}{9}$$

Answer 2

$$z^2=\dfrac{12}{9}+\dfrac{16}{9}i\\z^2=\dfrac{4}{9}(3+4i)\\z=\dfrac{2}{3}\sqrt{3+4i}$$ Let $a+bi=\sqrt{3+4i}$. Set a equation as: $$\begin{cases} a^2-b^2=3\\2ab=4 \end{cases}$$ From the second equation, we know $a=\dfrac{2}{b}$. Sub this into the first equation and you will solve the answer.