Complex number-finding roots

• Jul 31st 2006, 01:32 AM
kingkaisai2
Complex number-finding roots
Find the roots of the equation z^2=21-20i
• Jul 31st 2006, 04:30 AM
Soroban
Hello, kingkaisai2!

Edit: I corrected my wrong sign below ... Sorry for any confusion it caused.

I don't know what methods you've been taught . . .

Quote:

Find the roots of the equation $\displaystyle z^2 \:=\:21-20i$

Let $\displaystyle z = a + bi$, where $\displaystyle a$ and $\displaystyle b$ are real.

We have: .$\displaystyle (a + bi) \:= \:21 - 20i$

Then: .$\displaystyle a^2 + 2abi$+$\displaystyle b^2i^2\:=\:$$\displaystyle 21 - 20i\quad\Rightarrow\quad (a^2 - b^2) + 2abi \:=\:21 - 20i Equate real and imaginary components: .\displaystyle \begin{array}{cc}a^2 - b^2 \:= \:21\\ 2ab \:= \:-20\end{array} \displaystyle \begin{array}{cc}(1)\\(2)\end{array} From (2), we have: .\displaystyle b = -\frac{10}{a} Substitute into (1): .\displaystyle a^2 - \left(-\frac{10}{a}\right)^2\:=\:21\quad\Rightarrow\quad a^2 - \frac{100}{a^2}\:=\:21 Multiply by \displaystyle a^2:\;\;a^4 - 100\:=\:21a^2\quad\Rightarrow\quad a^4 - 21a^2 - 100\:=\:0 Factor: .\displaystyle (a^2 + 4)(a^2 - 25)\:=\:0 We have two equations to solve: . . \displaystyle a^2 + 4\:=\:0\quad\Rightarrow\quad a^2\,=\,-4\;\cdots \text{ but }a\text{ must be }real. . . \displaystyle a^2 - 25\:=\:0\quad\Rightarrow\quad a^2\,=$$\displaystyle \,25\quad\Rightarrow\quad a\,=\,\pm5$

Substitute into (2): .$\displaystyle b \:= \:-\frac{10}{\pm5} \:=\:\mp2$

Therefore: .$\displaystyle \boxed{z\;=\;\{5 - 2i,\;-5 + 2i\}}$

• Jul 31st 2006, 11:02 AM
Quick
Quote:

Originally Posted by Soroban
Hello, kingkaisai2!

I don't know what methods you've been taught . . .

Let $\displaystyle z = a + bi$, where $\displaystyle a$ and $\displaystyle b$ are real.

We have: .$\displaystyle (a + bi) \:= \:21 - 20i$

Then: .$\displaystyle a^2 + 2abi - b^2i^2\:=\:$$\displaystyle 21 - 20i\quad\Rightarrow\quad (a^2 - b^2) + 2abi \:=\:21 - 20i$

shouldn't it be $\displaystyle (a^2+b^2)+2abi=21-20i$? because i^2 equals -1