Complex number-finding roots

**kingkaisai2** · July 31st 2006, 01:32 AM

Find the roots of the equation z^2=21-20i

**Soroban** · July 31st 2006, 04:30 AM

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 . . .

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

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

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

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

Equate real and imaginary components: . $\begin{array}{cc}a^2 - b^2 \:= \:21\\ 2ab \:= \:-20\end{array}$ $\begin{array}{cc}(1)\\(2)\end{array}$

From (2), we have: . $b = -\frac{10}{a}$

Substitute into (1): . $a^2 - \left(-\frac{10}{a}\right)^2\:=\:21\quad\Rightarrow\quad a^2 - \frac{100}{a^2}\:=\:21$

Multiply by $a^2:\;\;a^4 - 100\:=\:21a^2\quad\Rightarrow\quad a^4 - 21a^2 - 100\:=\:0$

Factor: . $(a^2 + 4)(a^2 - 25)\:=\:0$

We have two equations to solve:

. . $a^2 + 4\:=\:0\quad\Rightarrow\quad a^2\,=\,-4\;\cdots \text{ but }a\text{ must be }real.$

. . $a^2 - 25\:=\:0\quad\Rightarrow\quad a^2\,=$ $\,25\quad\Rightarrow\quad a\,=\,\pm5$

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

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

**Quick** · July 31st 2006, 11:02 AM

Originally Posted by Soroban

Hello, kingkaisai2!

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

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

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

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

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

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