finding a solution for $m$ given $(1+i)z^2-2mz+m-2=0$

Question

Given the equation: $(1+i)z^2-2mz+m-2=0$, while $z$ is complex and $m$ is a parameter.

For which values of $m$ the equation has one solution?

So my idea was to use: $b^2-4ac=0$ for $ax^2+bx+c=0$

But it leads to difficult computation which i could not solve.

Is there any other way? or any way to solve this question?

Thanks.

This is a sound idea. Where did you have problems in the computation? — J.R., Jan 26 '14 at 13:44

J.R. · Accepted Answer · 2014-01-27T16:23:34.397

0

Your idea makes sense. The equation has one solution if and only if the discriminant $b^2-4ac$ is zero.

$$b^2-4ac=(-2m)^2-4(m-2)(1+i)=4m^2-(4+4i)m+(8+8i)$$

This is a quadratic equation in $m$ which has the two solutions $2i$ and $1-i$.

edited Jan 27 '14 at 16:23

answered Jan 26 '14 at 13:52

J.R.

17,904

i had a typo please see my edit – davon Jan 26 '14 at 17:39
@davon I changed my answer according to your edit. – J.R. Jan 27 '14 at 16:23

score 0 · Answer 2 · answered Jan 26 '14 at 14:30

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You should write your equation as: $(1+i)\cdot z^2 + (-2 m -1)\cdot z + m$

now you can use your idea: $b^2-4ac$ in your case: $(-2 m-1)^2-4(1+i)(m) = 4m^2 + 4m + 1 - 4m - 4mi = 4m^2 - 4mi + 1 = 0$ Solve this formula for quadratic equation leads to $ m_{1/2} = \frac{4i\pm\sqrt{-32}}{8} = \frac{4i\pm4\sqrt{2}i}{8} = \frac{1}{2}i \pm \frac{\sqrt{2}}{2}i$

answered Jan 26 '14 at 14:30

user114224

138

thanks but i had a typo it's: $(1+i)z^2-2mz+m-2=0$ – davon Jan 26 '14 at 17:41

finding a solution for $m$ given $(1+i)z^2-2mz+m-2=0$

2 Answers2