Writing a complex number in a+bi form

**Egoyan** · February 19th 2013, 07:13 AM

Hi guys,

I'm asked to write the number 5/i in the a+bi form. The format is such that I can only enter the values of a and b in the computer. I'm not sure what I should write. Can anyone help me please?

Thanks a lot!

Egoyan

**Plato** · February 19th 2013, 07:25 AM

Originally Posted by Egoyan

Hi guys,

I'm asked to write the number 5/i in the a+bi form. The format is such that I can only enter the values of a and b in the computer. I'm not sure what I should write. Can anyone help me please?

Learn this fact, it will save your life in complex numbers.
$(\forall z\in\mathbb{C}\setminus\{0\})\left[\frac{1}{z}=\frac{\overline{z}}{|z|^2}\right]$ .

So $\frac{1}{-3+4i}=\frac{-3-4i}{5}$ .

This $\frac{5}{i}=\frac{-5i}{1}=-5i$ .

**Egoyan** · February 19th 2013, 07:28 AM

Oh my. I did not know that. Heh. Thanks a lot! I won't forget it any time soon now...

**HallsofIvy** · February 19th 2013, 07:37 AM

More generally, a complex number, a+ bi, multiplied by its "complex conjugate", a- bi, gives a non-negative real number: $(a+ bi)(a- bi)= a^2+ abi - abi- b^2i^2= a^2+ b^2= |a|$ . In particular, you can make denominator or a fraction a real number by multiplying the numerator and denominator by the complex conjugate of the denominator:
$\frac{a+ bi}{c+ di}= \frac{a+ bi}{c+ di}\frac{c- di}{c- di}= \frac{(ac+ bd)+ i(bc- ad)}{c^2+ d^2}= \frac{ac+ bd}{c^2+ d^2}+ \frac{bc- ad}{c^2+ d^2}i$ .

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