Find all complex numbers z such that z^2=2i

please answer in a+bi

thank you​

2 Answers:

z = 1 + i   and   z = -1 - i

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Explanation:

We want z to be a complex number in the form z = a+bi, where a,b are real numbers and is imaginary.

Let's plug that into the equation your teacher gave you

You could use the FOIL rule to take a shortcut. I'm deciding to be a bit more wordy to show a further breakdown how everything is multiplying out.

Notice that the real part a^2-b^2 must be 0 so that it matches the real part on the right hand side.

a^2-b^2 = 0

(a-b)(a+b) = 0 .... difference of squares rule

a-b = 0 or a+b = 0

a = b or a = -b

So whatever solution z = a+bi is, it must have either a = b or a = -b.

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If a = b, then the 2abi portion on the left side turns into 2a^2*i

Set this equal to 2i on the right hand side and isolate 'a'

So a = 1 leads to b = 1

Or a = -1 leads to b = -1

Two complex solutions so far are:   z = 1 + i   and   z = -1 - i  based on those two cases above.

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Now consider the case that a = -b

We'll effectively have the same steps as the previous section, but the equation to solve now is

The only difference is that negative is out front. You should find that it leads to a^2 = -1, but this has no solutions because we stated earlier that a,b were real numbers.

So if a = -b, then it concludes with a,b being nonreal numbers. Ultimately we rule out the case that a = -b is possible.

Put another way, note how -2a^2 is always negative which clashes with the idea that the right hand side is positive (ignore the 'i' portions). This contradiction means that no real values of 'a' will make the equation to be true.

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To wrap things up, we only have two solutions and they are  

z = 1 + i   and   z = -1 - i

You can use a tool like WolframAlpha to confirm this.