Question

Find all complex numbers z such that z-=-32i, and give your answer in the form a+bi. Use the square root symbol 'V' where needed to give an exact value for your answer. z = ???

Answer 1

Given,

$z^{2}=-32i$

Let $z=a+bi$

$z^{2}=a^{2}-b^{2}+2abi=-32i$

Therefore,

Also we have,

$\left | z \right |^{2}=\left | z^{2} \right |$

$a^{2}+b^{2}=\left | -32i \right |$

$a^{2}+b^{2}=32$ ....................(3)

Solving (1) and (3)

$a^{2}=16$

$a=\pm 4$

$b=\pm 4$

However, from (2) we come to know that both a and b cannot be negative at the same time

Therefore, (a,b) has 2 solutions (4,-4) and (-4,4)

Therefore solution is,

Z= 4-4i

or, Z= -4+4i