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# imaginary numbers

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z+iz= 3+7i How do you get z?

Guest Dec 14, 2015

#1
+10

$$\displaystyle z(1+i) = 3 + 7i$$ ,

so

$$\displaystyle z = \frac{3+7i}{1+i}$$ .

Multiply top and bottom by the conjugate of the denominator.

Guest Dec 14, 2015
Sort:

#1
+10

$$\displaystyle z(1+i) = 3 + 7i$$ ,

so

$$\displaystyle z = \frac{3+7i}{1+i}$$ .

Multiply top and bottom by the conjugate of the denominator.

Guest Dec 14, 2015
#2
0

z=5+2i

Guest Dec 14, 2015
#4
+18948
+10

z+iz= 3+7i How do you get z?

$$\begin{array}{lrcl} & z+iz &=& 3+7i \\ & z(1+i) &=& 3+7i \\\\ \text{We set } & z = a+bi \\ & (a+bi)\cdot (1+i) &=& 3+7i \\ & a + ai+bi+bi^2 &=& 3+7i \qquad i^2 = -1\\ & a + ai+bi -b &=& 3+7i \\ & (a -b)+ (a+b) i &=& 3+7i \\\\ \text{We compare } &(1)\quad (a -b) &=& 3 \\ &(2)\quad (a+b) &=& 7 \\\\ & (a-b)+(a+b) &=& 3+7 \\ & a-b+a+b &=& 10 \\ & 2a &=& 10 \\ & \mathbf{a} & \mathbf{=} & \mathbf{5} \\\\ & (a+b)-(a-b) &=& 7-3 \\ & a+b-a+b &=& 4 \\ & 2b &=& 4 \\ & \mathbf{b} & \mathbf{=} & \mathbf{2} \\\\ & \mathbf{ z = a+bi }&\mathbf{=}& \mathbf{5+2i} \end{array}$$

heureka  Dec 14, 2015

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